On Two Open Problems on Double Vertex-Edge Domination in Graphs

Abstract

A vertex v of a graph $G=(V,E)$, ve-dominates every edge incident to v, as well as every edge adjacent to these incident edges. A set $S\subseteq V$ is a double vertex-edge dominating set if every edge of E is ve-dominated by at least two vertices of S. The double vertex-edge domination number ${\gamma }_{dve}\left(G\right)$ is the minimum cardinality of a double vertex-edge dominating set in G. A subset $S\subseteq V$ is a total dominating set (respectively, a 2-dominating set) if every vertex in V has a neighbor in S (respectively, every vertex in $V-S$ has at least two neighbors in S). The total domination number ${\gamma }_t\left(G\right)$ is the minimum cardinality of a total dominating set of G, and the 2-domination number ${\gamma }_2\left(G\right)$ is the minimum cardinality of a 2-dominating set of $G.$ Krishnakumari et al. (2017) showed that for every triangle-free graph $G,$${\gamma }_{dve}\left(G\right)\le {\gamma }_2\left(G\right)$, and in addition, if G has no isolated vertices, then ${\gamma }_{dve}\left(G\right)\le {\gamma }_t\left(G\right).$ Moreover, they posed the problem of characterizing those graphs attaining the equality in the previous bounds. In this paper, we characterize all trees T with ${\gamma }_{dve}\left(T\right)={\gamma }_t\left(T\right)$ or ${\gamma }_{dve}\left(T\right)={\gamma }_2\left(T\right)$.

1. Introduction

In this paper, G is a simple nontrivial connected graph with vertex set $V=V\left(G\right)$ and edge set $E=E\left(G\right)$. The order $\left|V\right|$ of G is denoted by $n=n\left(G\right)$. For a vertex $v\in V$, the “open neighborhood” of v is the set $N\left(v\right)=\{u\in V(G):uv\in E(G\left)\right\}$ and the “closed neighborhood” of v is the set $N\left[v\right]=N\left(v\right)\cup \left\{v\right\}$. The “degree” of a vertex $v\in V$ is ${\mathrm{deg}}_G\left(v\right)=\left|N\left(v\right)\right|$. A vertex of degree one is called a “pendant vertex” or a “leaf” and its neighbor is called a “support vertex”. A “strong support vertex” is a support vertex adjacent to at least two leaves and an “end support vertex” is a support vertex having at most one non-leaf neighbor. If v is a support vertex, then $L_v$ denotes the set of leaves adjacent to $v.$ A “pendant path” P of a graph G is an induced path starting from a leaf and containing only vertices of degree two in G as inner vertices. The “distance” between two vertices u and v in a connected graph G is the length of a shortest $uv$-path in $G.$ The “diameter” of connected graph G, denoted by $\mathrm{diam}\left(G\right)$, is the maximum value among minimum distances between all pairs of vertices of G. For a vertex v in a rooted tree T, let $C\left(v\right)$ and $D\left(v\right)$ denote the set of children and descendants of v, respectively, and let $D\left[v\right]=D\left(v\right)\cup \left\{v\right\}$. Also, the “depth” of v, $\mathrm{depth}\left(v\right)$, is the largest distance from v to a vertex in $D\left(v\right)$. The “maximal subtree” at v is the subtree of T induced by $D\left[v\right]$, and is denoted by $T_v$. We write $P_n$ for a path of order n. A “double star” $D{S}_{p,q}$ is a tree containing exactly two non-pendant vertices, one of which is adjacent to p leaves and the other is adjacent to q leaves. If $A\subseteq V\left(G\right)$ and f is a mapping from $V\left(G\right)$ into some set of numbers, then $f\left(A\right)={\sum }_{x\in A}f\left(x\right)$, and the sum $f\left(V\right(G\left)\right)$ is called the “weight” $\omega \left(f\right)$ of f.

A subset $S\subseteq V$ is a “2-dominating set”, abbreviated 2d-set, of G if every vertex in $V\backslash S$ is adjacent to at least two vertices in S and it is a “total dominating set” (td-set), if every vertex in V is adjacent to a vertex in S. The “2-domination number”, ${\gamma }_2\left(G\right)$ (respectively, “total domination number”, ${\gamma }_t\left(G\right)$) of G, is the minimum cardinality of a 2d-set (respectively, td-set) of G. A 2d-set (respectively, td-set) of G with minimum cardinality is called a ${\gamma }_2\left(G\right)$-set (respectively, ${\gamma }_t\left(G\right)$-set). The literature on the subject of domination and total domination in graphs has been surveyed and detailed in two books [1,2].

A vertex v is said to “ve-dominate” every edge incident to any vertex in $N\left[v\right]$. A set $S\subseteq V$ is a “vertex-edge dominating set” (or simply, a ve-dominating set) if for every edge $e\in E$, there exists a vertex $v\in S$ that ve-dominates e. The minimum cardinality of a ve-dominating set of G, ${\gamma }_{ve}\left(G\right)$, is called the “vertex-edge domination number”. Vertex-edge domination was introduced by Peters [3] in his 1986 PhD thesis and studied further in [4,5,6,7,8,9,10].

A vertex-edge dominating set $D\subseteq V$ is called a “double vertex-edge dominating set (or simply, a dve-dominating set) of G, if every edge of E is ve-dominated by at least two vertices of D. The “double vertex-edge domination number” of G, ${\gamma }_{dve}\left(G\right)$, is the minimum cardinality of a double ve-dominating set. The concept of double vertex-edge domination in graphs was introduced by Krishnakumari, Chellali, and Venkatakrishnan in [11], where they proved the following results.

Proposition 1

([11]). For every triangle-free graph G without isolated vertices,

$${\gamma }_{dve}\left(G\right)\le {\gamma }_t\left(G\right).$$

(1)

Proposition 2

([11]). For every graph G,

$${\gamma }_{dve}\left(G\right)\le {\gamma }_2\left(G\right).$$

(2)

Moreover, the authors [11] posed these two problems.

Problem 1.

Characterize all connected graphs G with${\gamma }_{dve}\left(G\right)={\gamma }_2\left(G\right)$.

Problem 2.

Characterize all nontrivial trees T with${\gamma }_{dve}\left(T\right)={\gamma }_t\left(T\right)$.

In this paper, we settle the above open problems for trees by providing a constructive characterization of all trees T with ${\gamma }_{dve}\left(T\right)={\gamma }_2\left(T\right)$ or ${\gamma }_{dve}\left(T\right)={\gamma }_t\left(T\right)$.

2. Preliminaries

In this section, we provide some definitions and observations that will be useful throughout the paper. We introduce some sets of vertices satisfying special condition.

Definition 1.

Let u be a vertex of a graph G. A subset S of vertices is said to be analmost dve-dominating setwith respect to u if the following conditions are fulfilled: (i) any edge not incident to u, isve-dominated by at least two vertices in S, and (ii) any edge incident to u is ve-dominated by at least one vertex in S. Define

$${\gamma }_{dve}(G;u)=\min \left\{\right|S|:Sisanalmostdve-dominatingsetwithrespecttou\}.$$

Clearly, any dve-dominating set on G is an almost dve-dominating set with respect to any vertex of G and so${\gamma }_{dve}(G;u)\le {\gamma }_{dve}\left(G\right)$for each $u\in V\left(G\right)$. Define

$$\begin{array}{ccc}{W}_G^1\hfill & =& \{u\in V\left(G\right)|{\gamma }_{dve}(G;u)={\gamma }_{dve}\left(G\right)\}.\hfill \end{array}$$

Definition 2.

For a graph G, define

$$\begin{array}{ccc}{W}_G^2\hfill & =\hfill & \{v\in V\left(G\right)\mid thereexistsa{\gamma }_{dve}\left(G\right)-setSsuchthat|S\cap N_G\left[v\right]|\ge 1\},\hfill \\ W_G^3\hfill & =\hfill & \{v\in V\left(G\right)\mid thereexistsa{\gamma }_2\left(G\right)-setSsuchthatv\in S\},\hfill \\ W_G^4\hfill & =\hfill & \{v\in V(G)\mid thereexistsnodve-dominatingsetSofGsuchthatv\in S\}.\hfill \end{array}$$

Observation 1.

All leaves of a graph G are in every 2-dominating set of $G.$

Observation 2.

For any connected graph G with diameter at least three, there exists a ${\gamma }_t\left(G\right)$-set (respectively, ${\gamma }_{dve}\left(G\right)$-set) that contains no leaf of G.

Proof. 

Among all ${\gamma }_t\left(G\right)$-sets, let D be one containing the minimum number of leaves of $G.$ Assume that there exists a leaf of G, say x with $x\in D.$ Let y be the support vertex adjacent to $x.$ Clearly, $y\in D$ (else x will be without neighbors in D), and $A=N\left(y\right)-L_v\ne \varnothing$ (as G has diameter at least three). If $A\cap D\ne \varnothing ,$ then $D-\left\{x\right\}$ is a td-set of $G,$ a contradiction. Therefore, $A\cap D=\varnothing .$ Then, replacing x in D by any vertex of A, provides a ${\gamma }_t\left(G\right)$-set containing fewer leaves than $D,$ contradicting our choice of $D.$

A similar argument can be used to prove also that G has a ${\gamma }_{dve}\left(G\right)$-set that contains no leaf. □

Proposition 3.

Let G be a nontrivial connected graph and $u\in V\left(G\right)$ a vertex of degree at least two which is a support vertex or adjacent to an end-support vertex. If $G^{\prime }$ is the graph obtained from G by adding a new vertex v connected by an edge to u, then ${\gamma }_{dve}\left(G\right)={\gamma }_{dve}\left(G^{\prime }\right)$ and ${\gamma }_t\left(G\right)={\gamma }_t\left(G^{\prime }\right)$.

Proof. 

If $G^{\prime }$ is a star, then the results are immediate. Suppose $G^{\prime }$ is not a star. Thus, both $G^{\prime }$ and G have diameter at least three. By Observation 2, any ${\gamma }_t\left(G\right)$-set (respectively, ${\gamma }_t\left(G^{\prime }\right)$-set) containing no leaves contains u, and so is a td-set of $G^{\prime }$ (respectively, G) yielding ${\gamma }_t\left(G^{\prime }\right)\le {\gamma }_t\left(G\right)$ (respectively, ${\gamma }_t\left(G\right)\le {\gamma }_t\left(G^{\prime }\right)$). Therefore, ${\gamma }_t\left(G^{\prime }\right)={\gamma }_t\left(G\right)$.

Assume now that D is a ${\gamma }_{dve}\left(G^{\prime }\right)$-set containing no leaves. Then to ve-dominate the edge $uv$, we may assume that $D\cap \left(N\right[u]-\{v\left\}\right)\ne \varnothing$ and D is clearly a dve-dominating set of G implying that ${\gamma }_{dve}\left(G^{\prime }\right)\ge {\gamma }_{dve}\left(G\right)$. On the other hand, clearly any ${\gamma }_{dve}\left(G\right)$-set containing no leaves is a double ve-dominating set of $G^{\prime }$ and so ${\gamma }_{dve}\left(G^{\prime }\right)\le {\gamma }_{dve}\left(G\right)$. Thus, ${\gamma }_{dve}\left(G^{\prime }\right)={\gamma }_{dve}\left(G\right)$. □

Proposition 4.

Let G be a nontrivial connected graph and $u\in V\left(G\right)$ such that $u{x}_1{x}_2$ is a path in G in which ${\mathrm{deg}}_G\left(x_1\right)=2$ and ${\mathrm{deg}}_G\left(x_2\right)=1$. If $G^{\prime }$ is a graph obtained from G by attaching a path $P_2=y_1{y}_2$ and joining $y_1$ to u, then ${\gamma }_t\left(G^{\prime }\right)={\gamma }_t\left(G\right)+1$, ${\gamma }_{dve}\left(G^{\prime }\right)={\gamma }_{dve}\left(G\right)+1$ and ${\gamma }_2\left(G^{\prime }\right)={\gamma }_2\left(G\right)+1$.

Proof. 

Clearly, any ${\gamma }_t\left(G\right)$-set (respectively, ${\gamma }_t\left(G^{\prime }\right)$-set) containing no leaves, contains $u,x_1$ (respectively, $u,x_1,y_1$) and for which adding (respectively, removing) vertex $y_1$ yields a td-set of $G^{\prime }$ (respectively, G). Therefore, ${\gamma }_t\left(G^{\prime }\right)\le {\gamma }_t\left(G\right)+1$ (respectively, ${\gamma }_t\left(G\right)\le {\gamma }_t\left(G^{\prime }\right)-1)$), and so ${\gamma }_t\left(G^{\prime }\right)={\gamma }_t\left(G\right)+1$. Similarly, we can see that ${\gamma }_{dve}\left(G^{\prime }\right)={\gamma }_{dve}\left(G\right)+1$ and ${\gamma }_2\left(G^{\prime }\right)={\gamma }_2\left(G\right)+1$. □

Proposition 5.

Let G be a nontrivial connected graph and let u be a vertex of G. If $G^{\prime }$ is the graph obtained from G by adding a path $P_4=x_1{x}_2{x}_3{x}_4$ and joining u to $x_1$, then ${\gamma }_t\left(G^{\prime }\right)={\gamma }_t\left(G\right)+2$ and ${\gamma }_{dve}\left(G^{\prime }\right)\ge {\gamma }_{dve}\left(G\right)+2$. Moreover, if $u\in W_G^2$, then ${\gamma }_{dve}\left(G^{\prime }\right)={\gamma }_{dve}\left(G\right)+2$.

Proof. 

Clearly, any ${\gamma }_t\left(G\right)$-set can be extended to a td-set of $G^{\prime }$ by adding $x_2,x_3$, and so ${\gamma }_t\left(G^{\prime }\right)\le {\gamma }_t\left(G\right)+2$.

To prove the inverse inequality, let $v\in N_G\left(u\right)$ and let D be a ${\gamma }_t\left(G^{\prime }\right)$-set containing no leaves. Then, $x_2,x_3\in D$. If $v\in D$ or $x_1\notin D$, then $D\backslash \{x_2,x_3\}$ is a td-set of G yielding ${\gamma }_t\left(G^{\prime }\right)\ge {\gamma }_t\left(G\right)+2$. Thus, assume that $v\notin D$ and $x_1\in D$. Then, $(D\backslash \{x_1,x_2,x_3\})\cup \left\{v\right\}$ is a total dominating set of G yielding ${\gamma }_t\left(G^{\prime }\right)\ge {\gamma }_t\left(G\right)+2$. Therefore, ${\gamma }_t\left(G^{\prime }\right)={\gamma }_t\left(G\right)+2$.

Similarly, we can see that ${\gamma }_{dve}\left(G^{\prime }\right)\ge {\gamma }_{dve}\left(G\right)+2$. Now, if $u\in W_G^2$, then clearly some ${\gamma }_{dve}\left(G\right)$-set contains a vertex $N_G\left[u\right]$, and so it can be extended to a dve-dominating set of $G^{\prime }$ by adding $x_2,x_3$ and thus ${\gamma }_{dve}\left(G^{\prime }\right)\le {\gamma }_{dve}\left(G\right)+2$. Therefore, ${\gamma }_{dve}\left(G^{\prime }\right)={\gamma }_{dve}\left(G\right)+2$ when $u\in W_G^2$. □

Proposition 6.

Let G be a nontrivial connected graph and let u be a vertex of G, such that $u{x}_1{x}_2$ is a pendant path in G with ${\mathrm{deg}}_G\left(x_2\right)=1$. If $G^{\prime }$ is the graph obtained from G by adding a path $P_3=y_1{y}_2{y}_3$ and joining u to $y_1$, then ${\gamma }_t\left(G^{\prime }\right)={\gamma }_t\left(G\right)+2$, ${\gamma }_2\left(G^{\prime }\right)={\gamma }_2\left(G\right)+2$ and ${\gamma }_{dve}\left(G^{\prime }\right)={\gamma }_{dve}\left(G\right)+2$.

Proof. 

Clearly, any ${\gamma }_t\left(G\right)$-set (respectively, ${\gamma }_t\left(G^{\prime }\right)$-set) D containing no leaves contains $u,x_1$ (respectively, $u,x_1,y_1,y_2$) and so $D\cup \{y_1,y_2\}$ (respectively, $D-\{y_1,y_2\}$) is a td-set of $G^{\prime }$ (respectively, G) yielding ${\gamma }_t\left(G^{\prime }\right)={\gamma }_t\left(G\right)+2$. Similarly, we can easily see that ${\gamma }_2\left(G^{\prime }\right)={\gamma }_2\left(G\right)+2$.

Assume now that D is a ${\gamma }_{dve}\left(G^{\prime }\right)$-set containing no leaves. Obviously, to dve-dominate the edges $x_1{x}_2,y_2{y}_3$, we must have $u,x_1,y_1,y_2\in D$ and thus $D-\{y_1,y_2\}$ is a dve-dominating of $G.$ Thus, ${\gamma }_{dve}\left(G^{\prime }\right)\ge {\gamma }_{dve}\left(G\right)+2$. The equality is obtained from the fact that any ${\gamma }_{dve}\left(G\right)$-set can be extended to a dve-dominating set of $G^{\prime }$ by adding $y_1,y_2$. □

Proposition 7.

Let G be a nontrivial connected graph and let u be a vertex of G such that $u{x}_1{x}_2{x}_3$ is a pendant path in G with ${\mathrm{deg}}_G\left(x_3\right)=1$. If $G^{\prime }$ is the graph obtained from G by adding a path $P_3=y_1{y}_2{y}_3$ and joining u to $y_1$, then ${\gamma }_t\left(G^{\prime }\right)={\gamma }_t\left(G\right)+2$ and ${\gamma }_{dve}\left(G^{\prime }\right)\le {\gamma }_{dve}\left(G\right)+2$. Furthermore, if $u\in W_G^1$, then ${\gamma }_{dve}\left(G^{\prime }\right)={\gamma }_{dve}\left(G\right)+2$.

Proof. 

As in the proof of Proposition 6, we can see that ${\gamma }_t\left(G^{\prime }\right)={\gamma }_t\left(G\right)+2$. The inequality ${\gamma }_{dve}\left(G^{\prime }\right)\le {\gamma }_{dve}\left(G\right)+2$ follows from the fact that any ${\gamma }_{dve}\left(G\right)$-set can be extended to a dve-dominating set of $G^{\prime }$ by adding $y_1,y_2$. Assume now that $u\in W_G^1$ and let D be a ${\gamma }_{dve}\left(G^{\prime }\right)$-set containing no leaves. Then we must have $x_1,x_2,y_1,y_2\in D$, and thus $D-\{y_1,y_2\}$ is an almost dve-dominating set of G with respect to $u.$ We deduce from $u\in W_G^1$ that $|D-\{y_1,y_2\}|\ge {\gamma }_{dve}(G;u)={\gamma }_{dve}\left(G\right)$ and so ${\gamma }_{dve}\left(G^{\prime }\right)\ge {\gamma }_{dve}\left(G\right)+2$. Hence ${\gamma }_{dve}\left(G^{\prime }\right)={\gamma }_{dve}\left(G\right)+2$ when $u\in W_G^1$. □

Let $k\ge 2$ be an integer and $H_k$ be the graph obtained from the complete bipartite graph $K_{1,k}$ (also known by a star) by subdividing every edge twice (see Figure 1).

Proposition 8.

Let G be a nontrivial connected graph and let u be a vertex of G. If $G^{\prime }$ is the graph obtained from G by adding the graph $H_k$ and joining u to the center vertex of $H_k$. Then, ${\gamma }_2\left(G^{\prime }\right)={\gamma }_2\left(G\right)+2k$ and ${\gamma }_{dve}\left(G^{\prime }\right)={\gamma }_{dve}\left(G\right)+2k$.

Proof. 

Clearly, any ${\gamma }_2\left(G\right)$-set can be extended to a 2d-set of $G^{\prime }$ by adding $a_i,c_i$ for $1\le i\le k$, implying that ${\gamma }_2\left(G^{\prime }\right)\le {\gamma }_2\left(G\right)+2k$.

To prove the inverse inequality, let D be ${\gamma }_2\left(G^{\prime }\right)$-set. Obviously, $c_i\in D$ for each i. Now to 2-dominate the vertex $b_i$ we must have $\left|D\cap \{a_i,b_i\}\right|\ge 1$ for each i. If $a\notin D$, then $D-V\left(H_k\right)$ is a 2d-set of G and if $a\in D$, then $(D-V\left(H_k\right))\cup \left\{u\right\}$ is a 2d-set of $G.$ In either case, ${\gamma }_2\left(G^{\prime }\right)\ge {\gamma }_2\left(G\right)+2k$, and thus ${\gamma }_2\left(G^{\prime }\right)={\gamma }_2\left(G\right)+2k$. Similarly, we can easily see that ${\gamma }_{dve}\left(G^{\prime }\right)={\gamma }_{dve}\left(G\right)+2k$. □

Observation 3.

Let G be a triangle-free graph without isolated vertices, and let H be a subgraph of G.

1. 

If ${\gamma }_{dve}\left(H\right)={\gamma }_t\left(H\right)$, ${\gamma }_t\left(G\right)\le {\gamma }_t\left(H\right)+s$ and ${\gamma }_{dve}\left(G\right)\ge {\gamma }_{dve}\left(H\right)+s$ for some non-negative integer s, then ${\gamma }_{dve}\left(G\right)={\gamma }_t\left(G\right)$.

2. 

If ${\gamma }_{dve}\left(G\right)={\gamma }_t\left(G\right)$, ${\gamma }_{dve}\left(G\right)\le {\gamma }_{dve}\left(H\right)+s$ and ${\gamma }_t\left(G\right)\ge {\gamma }_t\left(H\right)+s$ for some non-negative integer s, then ${\gamma }_{dve}\left(H\right)={\gamma }_t\left(H\right)$.

Proof. 

• We conclude from the assumptions that ${\gamma }_{dve}\left(G\right)\ge {\gamma }_{dve}\left(H\right)+s={\gamma }_t\left(H\right)+s\ge {\gamma }_t\left(G\right)$ and this leads to the result in (1).
• By inequality (1) and the assumptions, we have

  $${\gamma }_t\left(G\right)={\gamma }_{dve}\left(G\right)\le {\gamma }_{dve}\left(H\right)+s\le {\gamma }_t\left(H\right)+s\le {\gamma }_t\left(G\right).$$
  Thus, all inequalities occurring in the above chain become equalities. In particular, ${\gamma }_{dve}\left(H\right)={\gamma }_t\left(H\right)$.

□

The proof of next result is similar to the proof of Observation 3, and therefore we omit the details.

Observation 4.

Let H be a subgraph of a graph G.

1. 

If ${\gamma }_2\left(H\right)={\gamma }_{dve}\left(H\right)$, ${\gamma }_2\left(G\right)\le {\gamma }_2\left(H\right)+s$ and ${\gamma }_{dve}\left(G\right)\ge {\gamma }_{dve}\left(H\right)+s$ for some non-negative integer s, then ${\gamma }_2\left(G\right)={\gamma }_{dve}\left(G\right)$.

2. 

If ${\gamma }_2\left(G\right)={\gamma }_{dve}\left(G\right)$, ${\gamma }_{dve}\left(G\right)\le {\gamma }_{dve}\left(H\right)+s$ and ${\gamma }_2\left(G\right)\ge {\gamma }_2\left(H\right)+s$ for some non-negative integer s, then ${\gamma }_2\left(H\right)={\gamma }_{dve}\left(H\right)$.

We close this section with the following simple observation.

Observation 5.

If T is a tree of order $n\ge 4$ with $2\le \mathrm{diam}\left(T\right)\le 3$, then ${\gamma }_t\left(T\right)={\gamma }_{dve}\left(T\right)=2$ and ${\gamma }_{dve}\left(T\right)\ne {\gamma }_2\left(T\right)$.

3. Trees $\mathit{T}$ with ${\mathit{\gamma }}_{\mathit{dve}}\left(\mathit{T}\right)={\mathit{\gamma }}_{\mathbf{2}}\left(\mathit{T}\right)$

In this section, we provide a constructive characterization of all trees T with ${\gamma }_{dve}\left(T\right)={\gamma }_2\left(T\right)$. For this purpose, we define the family $\mathcal{F}$ of unlabeled trees T that can be obtained from a sequence $T_1,T_2,\dots ,$$T_k$$(k\ge 1)$ of trees such that $T_1\in \{P_2,P_3\}$ and $T=T_k$. If $k\ge 2$, $T_{i+1}$ can be obtained recursively from $T_i$ by one of the following operations.

• Operation ${\mathcal{T}}_1$: If $u\in V\left(T_i\right)$ and there is a pendant path $u{x}_1{x}_2$ such that ${\mathrm{deg}}_{T_i}\left(x_1\right)=2$ and ${\mathrm{deg}}_{T_i}\left(x_2\right)=1$, then ${\mathcal{T}}_1$ adds a path $P_2=y_1{y}_2$ and an edge $u{y}_1$ to obtain $T_{i+1}$.
• Operation ${\mathcal{T}}_2$: If $u\in V\left(T_i\right)$ and there is a pendant path $u{x}_1{x}_2$ such that ${\mathrm{deg}}_{T_i}\left(x_1\right)=2$ and ${\mathrm{deg}}_{T_i}\left(x_2\right)=1$, then ${\mathcal{T}}_2$ adds a path $P_3=y_1{y}_2{y}_3$ and an edge $u{y}_1$ to obtain $T_{i+1}$.
• Operation ${\mathcal{T}}_3$: If $u\in V\left(T_i\right)$, then ${\mathcal{T}}_3$ adds a copy of the graph $H_k$ for some $k\ge 2,$ connected by an edge from its center vertex a to u to obtain $T_{i+1}$.
• Operation ${\mathcal{T}}_4$: If $u\in W_{T_i}^1\cap W_{T_i}^4$, then ${\mathcal{T}}_4$ adds a path $P_5=x_1{x}_2{x}_3{x}_4{x}_5$ and an edge $u{x}_2$ to obtain $T_{i+1}$.
• Operation ${\mathcal{T}}_5$: If $u\in W_{T_i}^3$, then ${\mathcal{T}}_5$ adds a path $P_4=x_1{x}_2{x}_3{x}_4$ and an edge $u{x}_1$ to obtain $T_{i+1}$.
• Operation ${\mathcal{T}}_6$: If $u\in W_{T_i}^4$ and there is a pendant path $u{x}_1{x}_2{x}_3,$ then ${\mathcal{T}}_6$ adds a path $P_3=y_{1}y{y}_2$ and an edge $uy$ to obtain $T_{i+1}$.

Lemma 1.

If $T_i$ is a tree with ${\gamma }_{dve}\left(T_i\right)={\gamma }_2\left(T_i\right)$ and $T_{i+1}$ is a tree obtained from $T_i$ by one of the operations ${\mathcal{T}}_i(1\le i\le 6)$, then ${\gamma }_{dve}\left(T_{i+1}\right)={\gamma }_2\left(T_{i+1}\right)$.

Proof. 

If $T_{i+1}$ is a tree obtained from $T_i$ by Operation ${\mathcal{T}}_i(1\le i\le 3)$, then the result follows by Propositions 4, 6, and 8.

Let $T_{i+1}$ be a tree obtained from $T_i$ by Operation ${\mathcal{T}}_4$. Clearly, any ${\gamma }_2\left(T_i\right)$-set can be extended to a 2d-set of $T_{i+1}$ by adding $x_1,x_3,x_5$ and so ${\gamma }_2\left(T_{i+1}\right)\le {\gamma }_2\left(T_i\right)+3$. Now, let D be a ${\gamma }_{dve}\left(T_{i+1}\right)$-set containing no leaves. Clearly, to dve-dominate the edge $x_4{x}_5$, we must have $x_3,x_4\in D$. If $D\cap \{x_1,x_2\}=\varnothing$, then to dve-dominate the edge $x_1{x}_2$ we must have $u\in D$ and as $u\in W_{T_i}^4$, we have $|D\backslash \{x_3,x_4\}|\ge {\gamma }_{dve}\left(T_i\right)+1$. If $D\cap \{x_1,x_2\}\ne \varnothing$, then $(D\backslash \{x_1,x_2,x_3,x_4\})\cup \left\{u\right\}$ is an almost dve-dominating set of $T_i$ and, as above, we have $|(D\backslash \{x_1,x_2,x_3,x_4\})\cup \left\{u\right\}|\ge {\gamma }_{dve}\left(T_i\right)+1$. In either case, ${\gamma }_{dve}\left(T_{i+1}\right)\ge {\gamma }_{dve}\left(T_i\right)+3$. Therefore we conclude from Observation 4 (Item 1) that ${\gamma }_2\left(T_{i+1}\right)={\gamma }_{dve}\left(T_{i+1}\right)$.

Next, let $T_{i+1}$ be a tree obtained from $T_i$ by Operation ${\mathcal{T}}_5$. As $u\in W_{T_i}^3$, any ${\gamma }_2\left(T_i\right)$-set containing u can be extended to a 2d-set of $T_{i+1}$ by adding $x_2,x_4$, and thus ${\gamma }_2\left(T_{i+1}\right)\le {\gamma }_2\left(T_i\right)+2$. On the other hand, by Proposition 5, we have ${\gamma }_{dve}\left(T_{i+1}\right)\ge {\gamma }_{dve}\left(T_i\right)+2$. Therefore, by Observation 4 (Item 1), we obtain ${\gamma }_2\left(T_{i+1}\right)={\gamma }_{dve}\left(T_{i+1}\right)$.

Finally, let $T_{i+1}$ be a tree obtained from $T_i$ by Operation ${\mathcal{T}}_6$. Clearly, any ${\gamma }_2\left(T_i\right)$-set can be extended to a 2d-set of $T_{i+1}$ by adding $y_1,y_2$, and so ${\gamma }_2\left(T_{i+1}\right)\le {\gamma }_2\left(T_i\right)+2$. On the other hand, let D be a ${\gamma }_{dve}\left(T_{i+1}\right)$-set containing no leaves. Clearly, to dve-dominate the edges $y{y}_1,y{y}_2$ and $x_2{x}_3$, we must have $u,y,x_1,x_2\in D$. Then, $D-\left\{y\right\}$ is a dve-dominating set of $T^{\prime }$. As $u\in W_{T_i}^4$, we deduce that $|D-\left\{y\right\}|\ge {\gamma }_{dve}\left(T_i\right)+1$. Therefore, ${\gamma }_{dve}\left(T_{i+1}\right)\ge {\gamma }_{dve}\left(T_i\right)+2.$ By Observation 4 (Item 1), we obtain ${\gamma }_2\left(T_{i+1}\right)={\gamma }_{dve}\left(T_{i+1}\right)$. □

Theorem 1.

If $T\in \mathcal{F}$, then ${\gamma }_2\left(T\right)={\gamma }_{dve}\left(T\right)$.

Proof. 

Let $T\in \mathcal{F}$. Then there exists a sequence of trees $T_1,T_2,\dots ,T_k$$(k\ge 1)$, such that $T_1\in \{P_2,P_3\}$ and $T=T_k.$ Moreover, if $k\ge 2$, then $T_{i+1}$ can be obtained from $T_i$ by one of the aforementioned operations. We proceed by induction on the number of operations used to construct T. If $k=1$, then $T=T_1\in \{P_2,P_3\}$ and clearly ${\gamma }_2\left(T\right)={\gamma }_{dve}\left(T\right)$. Let $k\ge 2$, and assume that the result holds for each tree $T\in \mathcal{F}$, which can be obtained from a sequence of operations of length $k-1.$ Let $T^{\prime }=T_{k-1}$. By the induction hypothesis, ${\gamma }_2\left(T^{\prime }\right)={\gamma }_{dve}\left(T^{\prime }\right)$. As $T=T_k$ is obtained by one of the Operations ${\mathcal{T}}_i$ ($i=1,2,3,4,5,6$) from $T^{\prime }$, we conclude from Lemma 1 that ${\gamma }_2\left(T\right)={\gamma }_{dve}\left(T\right)$. □

Before stating the main theorem of this section, we give the following two useful observations.

Observation 6.

In a connected nontrivial graph G, every 2-dominating set of G is a dve-dominating set.

Proof. 

Let S be a 2-dominating set of $G,$ and let $e=xy$ be any edge of $G.$ Clearly, if $\left|S\cap \{u,v\}\right|=2,$ then e is ve-dominated once by x and once by $y.$ Therefore, we assume, without loss of generality, that $x\in S.$ As x has at least two neighbors in $S,$ edge, e is then ve-dominated at least twice by vertices in $N\left(x\right)\cap S.$ In either case, e is ve-dominated twice by vertices of $S,$ and desired result follows. □

Observation 7.

Let T be a tree of order $n\ge 4$ with ${\gamma }_2\left(T\right)={\gamma }_{dve}\left(T\right).$ Then, every support vertex of T is adjacent to at most two leaves.

Proof. 

Let D be a ${\gamma }_2\left(T\right)$-set, and let v be a support vertex of T such that $\left|L_v\right|\ge 3.$ Clearly, $L_v\subseteq D,$ and for every $w\in L_v,$$D-\left\{w\right\}$ is a dve-dominating set of T. Therefore, ${\gamma }_{dve}\left(T\right)\le \left|D\right|-1={\gamma }_2\left(T\right)-1,$ contradicting the fact ${\gamma }_2\left(T\right)={\gamma }_{dve}\left(T\right)$. □

Theorem 2.

Let T be a tree of order $n\ge 2$. Then ${\gamma }_2\left(T\right)={\gamma }_{dve}\left(T\right)$ if and only if $T\in \mathcal{F}$.

Proof. 

By Theorem 1, we only need to prove necessity. Let T be a nontrivial tree of order n with ${\gamma }_2\left(T\right)={\gamma }_{dve}\left(T\right)$. We proceed by induction on n. If $n\le 3$, then $T\in \{P_2,P_3\}$ and $T\in \mathcal{F}$. Therefore, let $n\ge 4$. By Observation 5, we have $\mathrm{diam}\left(T\right)\ge 4,$ and thus $n\ge 5.$ Assume that the result holds for every nontrivial tree $T^{\prime }$ having an order less than n and satisfying ${\gamma }_2\left(T^{\prime }\right)={\gamma }_{dve}\left(T^{\prime }\right)$. Let T be a tree of order n with ${\gamma }_2\left(T\right)={\gamma }_{dve}\left(T\right)$. Let $v_1{v}_2\dots v_k(k\ge 5)$ be a diametral path in T, such that ${\mathrm{deg}}_T\left(v_2\right)$ is as large as possible. Among such paths, we choose one so that ${\mathrm{deg}}_T\left(v_3\right)$ is as large as possible and root T at $v_k$. By Observation 7, $2\le {\mathrm{deg}}_T\left(v_2\right)\le 3.$ We consider the following cases.

Case 1.${\mathrm{deg}}_T\left(v_2\right)=2$. By the choice of diametrical path, we may assume that any child of $v_3$ with depth one has degree 2. We distinguish the following situations.

Subcase 1.1.${\mathrm{deg}}_T\left(v_3\right)\ge 3$ and $v_3$ has a child z with depth 0.

Let $T^{\prime }=T-\left\{z\right\}$. Let D be a ${\gamma }_2\left(T\right)$-set. Clearly $v_1,z\in D$ and $|D\cap \{v_2,v_3\left\}\right|=1$. Without loss of generality, we may assume that $v_3\in D$. Therefore, $D\backslash \left\{z\right\}$ is a 2d-set of $T^{\prime }$, which implies that ${\gamma }_2\left(T\right)\ge {\gamma }_2\left(T^{\prime }\right)+1$. On the other hand, any ${\gamma }_{dve}\left(T^{\prime }\right)$-dominating set of $T^{\prime }$ containing no leaves is a dve-dominating set of T implying that ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)$. This leads to the contradiction ${\gamma }_2\left(T\right)-1\ge {\gamma }_2\left(T^{\prime }\right)\ge {\gamma }_{dve}\left(T^{\prime }\right)\ge {\gamma }_{dve}\left(T\right)={\gamma }_2\left(T\right).$

Subcase 1.2.${\mathrm{deg}}_T\left(v_3\right)\ge 3$ and $v_3$ has a child with depth 1.

Let $T^{\prime }=T-T_{v_2}$. As there is a ${\gamma }_2\left(T\right)$-set D containing $v_3$, $D\backslash \left\{v_1\right\}$ is a 2d-set of $T^{\prime }$ and so ${\gamma }_2\left(T\right)\ge {\gamma }_2\left(T^{\prime }\right)+1$. On the other hand, if S is a ${\gamma }_{dve}\left(T^{\prime }\right)$-set containing no leaves, then $v_3\in S$ and thus $S\cup \left\{v_2\right\}$ is a dve-dominating set of $T.$ Therefore, ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+1$. By Observation 4 (Item 2), we have ${\gamma }_2\left(T^{\prime }\right)={\gamma }_{dve}\left(T^{\prime }\right)$. It follows from the induction hypothesis that $T^{\prime }\in \mathcal{F}$. Therefore, $T\in \mathcal{F}$, as it can be obtained from $T^{\prime }$ by Operation ${\mathcal{T}}_1$.

Subcase 1.3.${\mathrm{deg}}_T\left(v_3\right)=2$ and $v_4$ has at least one child with depth 1 and degree 2.

Let $T^{\prime }=T-T_{v_3}$. By Proposition 6 and by the assumption ${\gamma }_2\left(T\right)={\gamma }_{dve}\left(T\right)$, we have ${\gamma }_2\left(T^{\prime }\right)={\gamma }_{dve}\left(T^{\prime }\right)$. By the induction hypothesis, we have $T^{\prime }\in \mathcal{T}$, and as T can be obtained from $T^{\prime }$ by Operation ${\mathcal{T}}_2$, we deduce that $T\in \mathcal{F}$.

Subcase 1.4.${\mathrm{deg}}_T\left(v_3\right)=2$ and $v_4$ has a child y with depth 1 and at least degree 3.

By Observation 7, ${\mathrm{deg}}_T\left(y\right)=3$. Let $L_y=\{y_1,y_2\}$ and $T^{\prime }=T-T_y$. Let D be a ${\gamma }_2\left(T\right)$-set. Clearly, $y_1,y_2\in D$. If $y\in D,$ then $(D\backslash \{y,y_1,y_2\})\cup \left\{v_4\right\}$ is a 2d-set of $T^{\prime }$ and if $y\notin D,$ then $D\backslash \{y_1,y_2\}$ is a 2d-set of $T^{\prime }$. In either case, we obtain ${\gamma }_2\left(T\right)\ge {\gamma }_2\left(T^{\prime }\right)+2$. Now, let S be a ${\gamma }_{dve}\left(T^{\prime }\right)$-set. If $v_4\in S,$ then $S\cup \left\{y\right\}$ is a dve-dominating set of T, which leads to a contradiction as ${\gamma }_2\left(T\right)\ge {\gamma }_2\left(T^{\prime }\right)+2\ge {\gamma }_{dve}\left(T^{\prime }\right)+2\ge {\gamma }_{dve}\left(T\right)-1+2={\gamma }_2\left(T\right)+1.$ Thus $v_4\notin S$ and so $v_4\in W_{T^{\prime }}^4$. On the other hand, as $S\cup \{v_4,y\}$ is a dve-dominating set of $T,$ we have ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+2$. By Observation 4 (Item 2), we obtain ${\gamma }_2\left(T^{\prime }\right)={\gamma }_{dve}\left(T^{\prime }\right)$, and so by the induction hypothesis, we have $T^{\prime }\in \mathcal{F}$. Therefore, $T\in \mathcal{F}$ because it can be obtained from $T^{\prime }$ by Operation ${\mathcal{T}}_6$.

Subcase 1.5.${\mathrm{deg}}_T\left(v_3\right)=2$ and $v_4$ has $\ell \ge 1$ children with depth 0 and $k\ge 2$ children with depth 2.

Let $T^{\prime }=T-T_{v_4}$. Assume that $v_4{x}_1^i{x}_2^i{x}_3^i(1\le i\le k)$ are pendant paths in T. By the choice of diametrical path, we have $\mathrm{deg}\left(x_1^i\right)=\mathrm{deg}\left(x_2^i\right)=2$ for each $i\in \{1,\dots ,k\}$. We may assume that $\{x_1^i,x_3^i\mid 1\le i\le k\}\cup L_{v_4}$ is a subset of any ${\gamma }_2\left(T\right)$-set. Let D be a ${\gamma }_2\left(T\right)$-set. If $v_4\in D$, then replace $v_4$ by $v_5$ in D. Therefore, we can assume that $v_4\notin D$. It follows that $D\backslash (\{x_1^i,x_3^i\mid 1\le i\le k\}\cup L_{v_4})$ is a 2d-set of $T^{\prime }$ and thus ${\gamma }_2\left(T\right)\ge {\gamma }_2\left(T^{\prime }\right)+2k+\ell$. On the other hand, any ${\gamma }_{dve}\left(T^{\prime }\right)$-set can be extended to a dve-dominating set of T by adding the vertices $x_1^i,x_2^i(1\le i\le k)$, which implies that ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+2k$. This leads to the following contradiction,

$${\gamma }_2\left(T\right)={\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+2k\le {\gamma }_2\left(T^{\prime }\right)+2k\le {\gamma }_2\left(T\right)-\ell .$$

Subcase 1.6.${\mathrm{deg}}_T\left(v_3\right)=2$ and $v_4$ has $\ell \ge 2$ children with depth 0 and $k=1$ children with depth 2.

Let $T^{\prime }=T-T_{v_4}$. As in the Subcase 1.5, we can see that ${\gamma }_2\left(T\right)\ge {\gamma }_2\left(T^{\prime }\right)+2+\ell$. Also, if S is a ${\gamma }_{dve}\left(T^{\prime }\right)$-set, then $S\cup \{v_2,v_3,v_4\}$ is a dve-dominating set of T, and thus ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+3.$ Therefore,

$${\gamma }_2\left(T\right)={\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+3\le {\gamma }_2\left(T^{\prime }\right)+3\le {\gamma }_2\left(T\right)-2-\ell +3,$$

a contradiction.

Subcase 1.7.${\mathrm{deg}}_T\left(v_3\right)=2$ and $v_4$ has no child with depth 0 and $k\ge 2$ children with depth 2.

Let $T^{\prime }=T-T_{v_4}$. Assume that $v_4{x}_1^i{x}_2^i{x}_3^i(1\le i\le k)$ are pendant paths in T. By the choice of diametrical path, we have $\mathrm{deg}\left(x_1^i\right)=\mathrm{deg}\left(x_2^i\right)=2$ for each i. Thus, $T_{v_4}$ is isomorphic to the graph $H_k.$ As in Subcase 1.5, we can see that ${\gamma }_2\left(T\right)\ge {\gamma }_2\left(T^{\prime }\right)+2k$. On the other hand, any ${\gamma }_{dve}\left(T^{\prime }\right)$-set can be extended to a dve-dominating set of T by adding $x_1^i,x_2^i$$(1\le i\le k)$ which implies that ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+2k$. By Observation 4 (Item 2), we have ${\gamma }_2\left(T^{\prime }\right)={\gamma }_{dve}\left(T^{\prime }\right)$. It follows from the induction hypothesis that $T^{\prime }\in \mathcal{F}$. Therefore, $T\in \mathcal{F}$ because it can be obtained from $T^{\prime }$ by Operation ${\mathcal{T}}_3$.

Subcase 1.8.${\mathrm{deg}}_T\left(v_3\right)=2$ and $v_4$ has exactly one child of depth 0 and one child of depth 2.

Let w be the leaf adjacent to $v_4$ and let $T^{\prime }=T-T_{v_4}$. Note that $T^{\prime }$ is nontrivial, for otherwise ${\gamma }_2\left(T\right)\ne {\gamma }_{dve}\left(T\right).$ As in Subcase 1.5, we can see that ${\gamma }_2\left(T\right)\ge {\gamma }_2\left(T^{\prime }\right)+3$. Now, let S be a ${\gamma }_{dve}\left(T^{\prime }\right)$-set. If $v_5\in S$, then $S\cup \{v_2,v_3\}$ is a dve-dominating set of T and we get the following contradiction,

$${\gamma }_2\left(T\right)={\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+2\le {\gamma }_2\left(T^{\prime }\right)+2\le {\gamma }_2\left(T\right)-3+2.$$

Therefore, $v_5\notin S$, and thus $v_5\in W_{T^{\prime }}^4$. Then, $S\cup \{v_2,v_3,v_4\}$ is a dve-dominating set of T, implying that ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+3$. By Observation 4 (item 2), we have ${\gamma }_2\left(T^{\prime }\right)={\gamma }_{dve}\left(T^{\prime }\right)$, and thus ${\gamma }_2\left(T\right)={\gamma }_2\left(T^{\prime }\right)+3$ and ${\gamma }_{dve}\left(T\right)={\gamma }_{dve}\left(T^{\prime }\right)+3$. We deduce from the induction hypothesis that $T^{\prime }\in \mathcal{T}$. Next, we need to prove that $v_5\in W_{T^{\prime }}^1$. Suppose, to the contrary, that $v_5\notin W_{T^{\prime }}^1$. Then, any ${\gamma }_{dve}(T^{\prime };v_5)$-set can be extended to a dve-dominating set of T by adding $v_2,v_3,v_4$, which leads to the contradiction

$${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}(T^{\prime };v_5)+3\le {\gamma }_{dve}\left(T^{\prime }\right)-1+3={\gamma }_{dve}\left(T^{\prime }\right)+2.$$

Therefore, $v_5\in W_{T^{\prime }}^1$, and thus $v_5\in W_{T^{\prime }}^1\cap W_{T^{\prime }}^4.$ Consequently, T can be obtained from $T^{\prime }$ by Operation ${\mathcal{T}}_4$, yielding $T\in \mathcal{F}$.

Subcase 1.9.${\mathrm{deg}}_T\left(v_3\right)=2$ and ${\mathrm{deg}}_T\left(v_4\right)=2$. We distinguish the following situations.

(i)

${\mathrm{deg}}_T\left(v_5\right)=2$ or $v_5$ has a child with depth 0 or depth 1 or depth 2.

Let $T^{\prime }=T-T_{v_4}$. As in Subcase 1.5, we have ${\gamma }_2\left(T\right)\ge {\gamma }_2\left(T^{\prime }\right)+2$. Now, let S be a ${\gamma }_{dve}\left(T^{\prime }\right)$-set containing no leaf. Clearly, $S\cap N_{T^{\prime }}\left[v_5\right]\ne \varnothing$, and thus $S\cup \{v_2,v_3\}$ is a dve-dominating set of T, which implies that ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+2$. By Observation 4 (Item 2) and the induction hypothesis, we obtain $T^{\prime }\in \mathcal{F}$. Next, we prove that $v_5\in W_{T^{\prime }}^3$. Suppose, to the contrary, that $v_5\notin W_{T^{\prime }}^3$, and let D be a ${\gamma }_2\left(T\right)$-set. Without loss of generality, we may assume that $v_1,v_3,v_5\in D$. Then, $D\backslash \{v_1,v_3\}$ is a 2d-set of $T^{\prime }$ containing $v_5$, implying that ${\gamma }_2\left(T\right)-2=\left|D\right|-2\ge {\gamma }_2\left(T^{\prime }\right)+1$. Therefore,

$${\gamma }_{dve}\left(T\right)={\gamma }_2\left(T\right)\ge {\gamma }_2\left(T^{\prime }\right)+3\ge {\gamma }_{dve}\left(T^{\prime }\right)+3\ge {\gamma }_{dve}\left(T\right)+1,$$

a contradiction. Therefore, $v_5\in W_{T^{\prime }}^3$. Therefore, $T\in \mathcal{F}$ because it is obtained from $T^{\prime }$ by Operation ${\mathcal{T}}_5$.

(ii)

${\mathrm{deg}}_T\left(v_5\right)\ge 3$ and all children of $v_5$ are of depth 3.

Let $v_5{y}_4{y}_3{y}_2{y}_1$ be a path in T such that $y_4\notin \{v_4,v_6\}$. By seeing the previous cases, we have ${\mathrm{deg}}_T\left(y_4\right)={\mathrm{deg}}_T\left(y_3\right)={\mathrm{deg}}_T\left(y_2\right)=2$. Let $T^{\prime }=T-T_{v_4}$. As the Subcase 1.5, we have ${\gamma }_2\left(T\right)\ge {\gamma }_2\left(T^{\prime }\right)+2$. Let S be a ${\gamma }_{dve}\left(T^{\prime }\right)$-set containing no leaf such that $|S\cap \{y_2,y_3,y_4\left\}\right|$ is as small as possible. Clearly $y_2,y_3\in D$ and by the choice of S, we have $v_4\notin S$. Therefore, $S\cup \{v_2,v_3\}$ is a dve-dominating set of T and thus ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+2$. As in Item (i), we obtain ${\gamma }_2\left(T^{\prime }\right)={\gamma }_{dve}\left(T^{\prime }\right)$, $T^{\prime }\in \mathcal{F}$, and $v_5\in W_{T^{\prime }}^3$. Therefore, $T\in \mathcal{F}$, because it is obtained from $T^{\prime }$ by Operation ${\mathcal{T}}_5$.

Case 2.${\mathrm{deg}}_T\left(v_2\right)=3$.

Assume that $L_{v_2}=\{v_1,w\}$. We consider the following subcases.

Subcase 2.1.$\mathrm{deg}\left(v_3\right)\ge 3$ and $v_3$ has a child with depth 1, say $y,$ with $y\ne v_2$.

Let $T^{\prime }=T-T_{v_2}$. Assume that D is a ${\gamma }_2\left(T\right)$-set. If $v_2\in D$, then $D\backslash \{v_1,v_2,w\}\cup \left\{v_3\right\}$ is a 2d-set of $T^{\prime }$, and if $v_2\notin D$, then $D\backslash \{v_1,w\}$ is a 2d-set of $T^{\prime }$; so, in either case, ${\gamma }_2\left(T\right)\ge {\gamma }_2\left(T^{\prime }\right)+2$. Now, let S be a ${\gamma }_{dve}\left(T^{\prime }\right)$-set containing no leaves. As $y,v_3\in S,$$S\cup \left\{v_2\right\}$ is a dve-dominating set of T, and thus ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+1.$, it follows that

$${\gamma }_2\left(T\right)={\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+1\le {\gamma }_2\left(T^{\prime }\right)+1\le {\gamma }_2\left(T\right)-1,$$

a contradiction.

Subcase 2.2.$\mathrm{deg}\left(v_3\right)\ge 3$ and $v_3$ has $\ell \ge 1$ children with depth 0.

Let D be a ${\gamma }_2\left(T\right)$-set. Then $\{v_1,w\}\cup$$L_{v_3}\subset D,$ and thus $\{v_2,v_3\}\cup D-(\{v_1,w\}\cup$$L_{v_3})$ is a dve-dominating set of T of cardinality at most ${\gamma }_2\left(T\right)-1,$ a contradiction.

Subcase 2.3.$\mathrm{deg}\left(v_3\right)=2.$

Let D be a ${\gamma }_2\left(T\right)$-set $D.$ Clearly, D contains $v_1,w.$ Moreover, to 2-dominate $v_3$, we must have $\{v_2,v_3\}\cap D\ne \varnothing ,$, say $v_3\in D.$ However, then, $\left\{v_2\right\}\cup D-\{v_1,w\}$ is a dve-dominating set of T of cardinality at most ${\gamma }_2\left(T\right)-1,$ a contradiction. □

4. Trees $\mathit{T}$ with ${\mathit{\gamma }}_{\mathit{dve}}\left(\mathit{T}\right)={\mathit{\gamma }}_{\mathit{t}}\left(\mathit{T}\right)$

In this section, we provide a constructive characterization of all trees T with ${\gamma }_{dve}\left(T\right)={\gamma }_t\left(T\right)$. For this purpose, we define the family $\mathcal{T}$ of unlabeled trees T that can be obtained from a sequence $T_1,T_2,\dots ,$$T_k$$(k\ge 1)$ of trees, such that $T_1\in \{P_2,P_3,P_4\}$ and $T=T_k$. If $k\ge 2$, then $T_{i+1}$ can be obtained recursively from $T_i$ by one of the following operations.

• Operation ${\mathcal{O}}_1$: If $u\in V\left(T_i\right)$ is a support vertex or a non-leaf vertex adjacent to an end support vertex, then ${\mathcal{O}}_1$ adds a vertex x and an edge $ux$ to obtain $T_{i+1}$.
• Operation ${\mathcal{O}}_2$: If $u\in V\left(T_i\right)$ and there is a path $u{x}_1{x}_2$ in $T_i$ such that ${\mathrm{deg}}_{T_i}\left(x_1\right)=2$ and ${\mathrm{deg}}_{T_i}\left(x_2\right)=1$, then ${\mathcal{O}}_2$ adds a path $P_2=y_1{y}_2$ and an edge $u{y}_1$ to obtain $T_{i+1}$.
• Operation ${\mathcal{O}}_3$: If $u\in W_{T_i}^2$, then ${\mathcal{O}}_3$ adds a path $P_4=x_1{x}_2{x}_3{x}_4$ and an edge $u{x}_1$ to obtain $T_{i+1}$.
• Operation ${\mathcal{O}}_4$: If $u\in V\left(T_i\right)$ and there is a path $u{x}_1{x}_2$ in $T_i$ such that ${\mathrm{deg}}_{T_i}\left(x_1\right)=2$ and ${\mathrm{deg}}_{T_i}\left(x_2\right)=1$, then ${\mathcal{O}}_4$ adds a path $P_3=y_1{y}_2{y}_3$ and an edge $u{y}_1$ to obtain $T_{i+1}$.
• Operation ${\mathcal{O}}_5$: If $u\in W_{T_i}^1$ and there is a path $u{x}_1{x}_2{x}_3$ in $T_i$ such that ${\mathrm{deg}}_{T_i}\left(x_1\right)={\mathrm{deg}}_{T_i}\left(x_2\right)=2$ and ${\mathrm{deg}}_{T_i}\left(x_3\right)=1$, then ${\mathcal{O}}_5$ adds a path $y_1{y}_2{y}_3$ and an edge $u{y}_1$ to obtain $T_{i+1}$.
• Operation ${\mathcal{O}}_6$: If $u\in V\left(T_i\right)$ is a leaf and there is a path $u{x}_1{x}_2{x}_3$ in $T_i$ such that ${\mathrm{deg}}_{T_i}\left(x_3\right)=1$ and no ${\gamma }_{dve}\left(T_i\right)$-set contains both u and $x_1$, then ${\mathcal{O}}_6$ adds a vertex x and an edge $ux$ to obtain $T_{i+1}$.
• Operation ${\mathcal{O}}_7$: If $u\in V\left(T_i\right)$ is a support vertex with $v\in L_u$ and there is a path $u{y}_2{y}_1$ such that ${\mathrm{deg}}_{T_i}\left(y_2\right)=2$ and ${\mathrm{deg}}_{T_i}\left(y_1\right)=1$, then ${\mathcal{O}}_7$ adds a new vertex w and a path $P_4=x_1{x}_2{x}_3{x}_4$ and edges $y_{1}w,$$v{x}_4$ to obtain $T_{i+1}$.

Lemma 2.

If $T_i$ is a tree with ${\gamma }_{dve}\left(T_i\right)={\gamma }_t\left(T_i\right)$ and $T_{i+1}$ is a tree obtained from $T_i$ by one of the operations ${\mathcal{O}}_j(1\le j\le 7)$, then ${\gamma }_{dve}\left(T_{i+1}\right)={\gamma }_t\left(T_{i+1}\right)$.

Proof. 

If $T_{i+1}$ is a tree obtained from $T_i$ by Operation ${\mathcal{O}}_j,$ where $j\in \{1,\dots ,5\}$, then the result follows by Propositions 3, 4, 5, 6, and 7.

Let $T_{i+1}$ be obtained from $T_i$ by Operation ${\mathcal{O}}_6$. Clearly, any ${\gamma }_t\left(T_i\right)$-set can be extended to a td-set of T by adding u which implies that ${\gamma }_t\left(T_{i+1}\right)\le {\gamma }_t\left(T_i\right)+1$. Moreover, if D is a ${\gamma }_{dve}\left(T_{i+1}\right)$-set containing no leaf, then clearly $u,x_1\in D$ and D is a double ve-dominating set of $T_i$. As, by the assumption in Operation ${\mathcal{O}}_6$, no ${\gamma }_{dve}\left(T_i\right)$-set contains both u and $x_1,$ we deduce that $\left|D\right|>{\gamma }_{dve}\left(T_i\right),$, that is, ${\gamma }_{dve}\left(T_{i+1}\right)\ge {\gamma }_{dve}\left(T_i\right)+1$. Now, by Observation 3 (Item 1), we obtain ${\gamma }_{dve}\left(T_{i+1}\right)={\gamma }_t\left(T_{i+1}\right)$.

Assume now that $T_{i+1}$ is obtained from $T_i$ by Operation ${\mathcal{O}}_7$. As any ${\gamma }_t\left(T_i\right)$-set can be extended to a td-set of T by adding $x_2,x_3,y_1,$, we have ${\gamma }_t\left(T_{i+1}\right)\le {\gamma }_t\left(T_i\right)+3$. Let D be a ${\gamma }_{dve}\left(T_{i+1}\right)$-set containing no leaf. Then, $y_1,y_2,x_2,x_3\in D$, and to dve-dominate the edge $v{x}_4$, we may assume that $u\in D$. Therefore, $D-\{y_1,x_2,x_3\}$ is a dve-dominating set of $T_i$, implying that ${\gamma }_{dve}\left(T_{i+1}\right)\ge {\gamma }_{dve}\left(T_i\right)+3$. By Observation 3 (Item 1), we obtain ${\gamma }_{dve}\left(T_{i+1}\right)={\gamma }_t\left(T_{i+1}\right)$. □

Now we are ready to state the main theorem of this section.

Theorem 3.

Let T be a tree of order $n\ge 2$. Then ${\gamma }_{dve}\left(T\right)={\gamma }_t\left(T\right)$ if and only if $T\in \mathcal{T}$.

Proof. 

Let $T\in \mathcal{T}$. Then there exists a sequence of trees $T_1,T_2,\dots ,T_k$$(k\ge 1)$, such that $T_1\in \{P_2,P_3,P_4\}$ and $T=T_k.$ Moreover, if $k\ge 2$, then $T_{i+1}$ can be obtained from $T_i$ by one of the aforementioned operations. We proceed by induction on the number of operations used to construct T. If $k=1$, then $T\in \{P_2,P_3,P_4\}$ and clearly ${\gamma }_{dve}\left(T\right)={\gamma }_t\left(T\right)$. Assume now that $k\ge 2$ and that the result holds for each tree, $T\in \mathcal{T}$, that can be obtained from a sequence of length $k-1$, and let $T^{\prime }=T_{k-1}$. By the induction hypothesis, ${\gamma }_{dve}\left(T^{\prime }\right)={\gamma }_t\left(T^{\prime }\right)$. As $T=T_k$ is obtained from $T^{\prime }$ by using one of the Operations ${\mathcal{O}}_j,$ where $j\in \{1,2,\dots ,7\}$, we conclude from the Lemma 2 that ${\gamma }_{dve}\left(T\right)={\gamma }_t\left(T\right)$.

Conversely, let ${\gamma }_{dve}\left(T\right)={\gamma }_t\left(T\right)$. We proceed by induction on n. If $n\le 3$, then $T\in \{P_2,P_3\}$ and clearly $T\in \mathcal{T}$. Let $n\ge 4$ and let the result hold for every tree $T^{\prime }$ of order less than n, satisfying ${\gamma }_{dve}\left(T^{\prime }\right)={\gamma }_t\left(T^{\prime }\right)$. Let T be a tree of order n with ${\gamma }_{dve}\left(T\right)={\gamma }_t\left(T\right)$. If $\mathrm{diam}\left(T\right)=2$, then T is a star, and it can be obtained from $P_3$ by frequently use of Operation ${\mathcal{O}}_1$, and so $T\in \mathcal{T}$. If $\mathrm{diam}\left(T\right)=3$, then T is a double star $D{S}_{p,q},(q\ge p\ge 1)$. If $q=1$, then $T=P_4\in \mathcal{T}$. If $q>1$, then T can be obtained from $P_4$ by frequently use of ${\mathcal{O}}_1$, and thus $T\in \mathcal{T}$. Henceforth, we assume that $\mathrm{diam}\left(T\right)\ge 4$. Let $v_1{v}_2\dots v_k(k\ge 5)$ be a diametral path in T such that ${\mathrm{deg}}_T\left(v_2\right)$ is as large as possible. Root T at $v_k$.

If ${\mathrm{deg}}_T\left(v_2\right)\ge 3$, then let $T^{\prime }=T-\left\{v_1\right\}$. By Proposition 3 and by the induction hypothesis, we have $T^{\prime }\in \mathcal{T}$. Now, T can be obtained from $T^{\prime }$ by operation ${\mathcal{O}}_1$ and so $T\in \mathcal{T}$. Therefore, we assume that ${\mathrm{deg}}_T\left(v_2\right)=2$. By the choice of diametrical path, we may assume that any child of $v_3$ with depth one is of degree two. Note that using an argument similar to the previous one, we may assume that T has no strong support vertex. We consider the following cases.

Case 1.${\mathrm{deg}}_T\left(v_3\right)\ge 3$.

If $v_3$ has at least one child with depth 0, say x, then let $T^{\prime }=T-\left\{x\right\}$. By Proposition 3 and by the induction hypothesis, we have $T^{\prime }\in \mathcal{T}$. It follows that $T\in \mathcal{T}$ because it is obtained from $T^{\prime }$ by operation ${\mathcal{O}}_1$. Therefore, we assume that $v_3$ is not a support vertex. Then, $v_3$ has at least one children with depth 1, say $z_2,$ different from $v_2$. Let $v_3{z}_2{z}_1$ be a pendant path in T. Suppose $T^{\prime }=T-\{v_1,v_2\}$. By Proposition 4 and by the induction hypothesis, we have $T^{\prime }\in \mathcal{T}$. Now, as T is obtained from $T^{\prime }$ by operation ${\mathcal{O}}_2,$, we obtain $T\in \mathcal{T}$.

Case 2.${\mathrm{deg}}_T\left(v_3\right)=2$.

First, let ${\mathrm{deg}}_T\left(v_4\right)\ge 3$. If $v_4$ has a child $y_2$ with depth 1, then let $v_4{y}_2{y}_1$ be a pendant path in T, and let $T^{\prime }=T-T_{v_3}$. By Proposition 6 and the inductive hypothesis on $T^{\prime }$, we have $T^{\prime }\in \mathcal{T}$. It follows that $T\in \mathcal{T}$, as it can be obtained from $T^{\prime }$ by operation ${\mathcal{O}}_4.$

Assume now that $v_4$ has a child $z_3$ with depth 2, different from $v_3.$ According to the previous case, the path $v_4{z}_3{z}_2{z}_1$, having $z_1$ as a leaf, is a pendant path in $T.$ Let $T^{\prime }=T-T_{v_3}$. By Proposition 7 and Observation 3 (item 2), we have ${\gamma }_{dve}\left(T^{\prime }\right)={\gamma }_t\left(T^{\prime }\right)$, ${\gamma }_t\left(T\right)={\gamma }_t\left(T^{\prime }\right)+2$ and ${\gamma }_{dve}\left(T\right)={\gamma }_{dve}\left(T^{\prime }\right)+2$. By the induction hypothesis on $T^{\prime }$, we have $T^{\prime }\in \mathcal{T}$. We prove now that $v_4\in W_{T^{\prime }}^1$. Suppose, to the contrary, that $v_4\notin W_{T^{\prime }}^1$. Then ${\gamma }_{dve}(T^{\prime };v_4)\le {\gamma }_{dve}\left(T^{\prime }\right)-1$ and any ${\gamma }_{dve}(T^{\prime };v_4)$-set can be extended to a dve-dominating set of T by adding $v_2,v_3$ implying that ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}(T^{\prime };v_4)+2\le {\gamma }_{dve}\left(T^{\prime }\right)-1+2={\gamma }_{dve}\left(T^{\prime }\right)+1$, a contradiction. Therefore, $v_4\in W_{T^{\prime }}^1$, and therefore $T\in \mathcal{T}$ because it can be obtained from $T^{\prime }$ by operation ${\mathcal{O}}_5.$

Finally, suppose $v_4$ is a support vertex, and let $T^{\prime }=T-\left\{v_1\right\}$. Clearly, for any ${\gamma }_t\left(T\right)$-set D containing no leaf, we have $v_2,v_3,v_4\in D$, and thus $D\backslash \left\{v_2\right\}$ is a td-set of $T^{\prime },$ implying that ${\gamma }_t\left(T\right)\ge {\gamma }_t\left(T^{\prime }\right)+1$. On the other hand, any ${\gamma }_{dve}\left(T^{\prime }\right)$-set containing no leaves, contains $v_3,v_4$ and thus such a set plus $v_2$ is a dve-dominating set of T, implying that ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+1$. By Observation 3 (Item 2), we have ${\gamma }_{dve}\left(T^{\prime }\right)={\gamma }_t\left(T^{\prime }\right)$. By the induction hypothesis, we have $T^{\prime }\in \mathcal{T}$. Note that Observation 3 (Item 2) also implies that ${\gamma }_{dve}\left(T\right)={\gamma }_{dve}\left(T^{\prime }\right)+1.$ This means that no ${\gamma }_{dve}\left(T^{\prime }\right)$-set S contains both $v_2$ and $v_3$; otherwise, S dve-dominates T too, which contradicts ${\gamma }_{dve}\left(T\right)={\gamma }_{dve}\left(T^{\prime }\right)+1.$ Now, as T can be obtained from $T^{\prime }$ by operation ${\mathcal{O}}_6,$ we conclude that $T\in \mathcal{T}$.

From now on, we can assume that ${\mathrm{deg}}_T\left(v_4\right)=2$. We distinguish the following subcases.

Subcase 2.1.${\mathrm{deg}}_T\left(v_5\right)\ge 3$.

Let $T^{\prime }=T-T_{v_4}$. By Proposition 5, we have ${\gamma }_t\left(T\right)={\gamma }_t\left(T^{\prime }\right)+2$. Let S be a ${\gamma }_{dve}\left(T^{\prime }\right)$-set containing no leaves. If $v_5$ is a support vertex or is adjacent to a support vertex, then clearly $|S\cap N_{T^{\prime }}\left[v_5\right]|\ge 1$. If there is a path $v_5{z}_3{z}_2{z}_1$ such that $z_3\notin \{v_4,v_6\}$ and $\mathrm{deg}\left(z_1\right)=1$, then $z_3\in S$, and thus $|S\cap N_{T^{\prime }}\left[v_5\right]|\ge 1$. If there is a path $v_5{z}_4{z}_3{z}_2{z}_1$ such that $z_4\notin \{v_4,v_6\}$ and $\mathrm{deg}\left(z_1\right)=1$, then, as $z_1$ belongs to diametral path of T, each of $z_4,z_3$ and $z_2$ has degree two, and thus $z_2,z_3\in S.$ However, to ve-dominate the edge $v_5{z}_4$ we need that $|S\cap N_{T^{\prime }}\left[v_5\right]|\ge 1$. In either case, we deduce that $v_5\in W_{T^{\prime }}^2$. In this case, $S\cup \{v_2,v_3\}$ is a dve-dominating set of $T,$ implying that ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+2$. By Observation 3 (Item 2), we obtain ${\gamma }_t\left(T^{\prime }\right)={\gamma }_{dve}\left(T^{\prime }\right)$, and it follows from the induction hypothesis that $T^{\prime }\in \mathcal{T}$. Therefore, $T\in \mathcal{T}$ because it is obtained from $T^{\prime }$ by Operation ${\mathcal{O}}_3$.

Subcase 2.1.${\mathrm{deg}}_T\left(v_5\right)=2$.

We note that if $v_6$ is a leaf, then $T=P_6$ that belongs to $\mathcal{T}$, as it can be obtained from a path $P_2$ (whose vertices are in $W_{P_2}^2$) by using Operation ${\mathcal{O}}_3$. Therefore, we assume that ${\mathrm{deg}}_T\left(v_6\right)\ge 2.$ We consider some further situations.

(a)

$v_6$ has a child with depth 0 or 1 or ${\mathrm{deg}}_T\left(v_6\right)=2$.

Let $T^{\prime }=T-T_{v_4}$. By Proposition 5, ${\gamma }_t\left(T\right)={\gamma }_t\left(T^{\prime }\right)+2$ and ${\gamma }_{dve}\left(T\right)\ge {\gamma }_{dve}\left(T^{\prime }\right)+2$. On the other hand, any ${\gamma }_{dve}\left(T^{\prime }\right)$-set containing no leaves can be extended to a dve-dominating set of T by adding two vertices, $v_2$ and $v_3$, which implies that ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+2$. By Observation 3 (Item 2), we obtain ${\gamma }_t\left(T^{\prime }\right)={\gamma }_{dve}\left(T^{\prime }\right)$, and it follows from the induction hypothesis that $T^{\prime }\in \mathcal{T}$. Also, by assumption, it is clear that $v_5\in W_{T^{\prime }}^2$, and therefore $T\in \mathcal{T}$, as it can be obtained from $T^{\prime }$ by Operation ${\mathcal{O}}_3$.

(b)

$v_6$ has a child with depth 3.

Let $y_4$ be a child of $v_6$ with depth 3 such that $v_6{y}_4{y}_3{y}_2{y}_1$ is a path in T. As T has no strong support vertex, we have $\mathrm{deg}\left(y_2\right)=2$. If $\mathrm{deg}\left(y_3\right)\ge 3$, then by a similar argument to that used in Case 1, we can easily see that $T\in \mathcal{T}$. Therefore, we may assume that $\mathrm{deg}\left(y_3\right)=2$. If $\mathrm{deg}\left(y_4\right)\ge 3$, then, again, by using a similar argument as in the beginning of Case 2, we have $T\in \mathcal{T}$. Thus, we may assume that $\mathrm{deg}\left(y_4\right)=2.$ Now, let $T^{\prime }=T-T_{y_4}$. By Proposition 5, we have ${\gamma }_t\left(T\right)={\gamma }_t\left(T^{\prime }\right)+2$. Let S be a ${\gamma }_{dve}\left(T^{\prime }\right)$-set containing no leaves. Then, $v_2,v_3\in S$, and to dve-dominate the edge $v_4{v}_5$, we must have $S\cap \{v_4,v_5,v_6\}\ne \varnothing ,$ say $v_6\in S.$ Hence $v_6\in W_{T^{\prime }}^2$. Also, $S\cup \{y_2,y_3\}$ is a dve-dominating set of T, and thus ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+2$. By Observation 3 (Item 2), we obtain ${\gamma }_t\left(T^{\prime }\right)={\gamma }_{dve}\left(T^{\prime }\right),$, and by the induction hypothesis, we obtain $T^{\prime }\in \mathcal{T}$. Therefore, $T\in \mathcal{T}$, as it is obtained from $T^{\prime }$ by Operation ${\mathcal{O}}_3$.

(c)

$v_6$ has a child with depth 4, different from $v_5$.

Let $y_5$ be a child of $v_6$ with depth 4 such that $v_6{y}_5{y}_4{y}_3{y}_2{y}_1$ is a path in T. Note that, as $y_1$ belongs to a diametral path of T, and seeing the above situations, we can assume that $v_6{y}_5{y}_4{y}_3{y}_2{y}_1$ is a pendant path. Let $T^{\prime }=T-T_{y_4}$. By Proposition 5, ${\gamma }_t\left(T\right)={\gamma }_t\left(T^{\prime }\right)+2$. Let S be a ${\gamma }_{dve}\left(T^{\prime }\right)$-set containing no leaves. As above, $v_2,v_3\in S$, and we may assume that $v_6\in S$ to double ve-dominating the edge $v_4{v}_5.$ Thus $y_5\in W_{T^{\prime }}^2$, and $S\cup \{y_2,y_3\}$ is a dve-dominating set of $T,$, implying that ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+2$. By Observation 3 (Item 2), we obtain ${\gamma }_t\left(T^{\prime }\right)={\gamma }_{dve}\left(T^{\prime }\right)$, and thus $T^{\prime }\in \mathcal{T}$. Now, as T is obtained from $T^{\prime }$ by Operation ${\mathcal{O}}_3$, we deduce that $T\in \mathcal{T}$.

(d)

$v_6$ has a child with depth 2.

Considering above situations, we may assume that any child of $v_6$ other than $v_5$ has depth 2. First, let ${\mathrm{deg}}_T\left(v_6\right)=3$. Let $y_3$ be the child of $v_6$ with depth 2 such that $v_6{y}_3{y}_2{y}_1$ is a path in T. As T has no strong support vertex we have ${\mathrm{deg}}_T\left(y_2\right)=2$. Also, if ${\mathrm{deg}}_T\left(y_3\right)\ge 3$, then, as in the Case 1, we can see that $T\in \mathcal{T}$. Therefore, we may assume that ${\mathrm{deg}}_T\left(y_3\right)=2$. Let $T^{\prime }$ be the tree obtained from T by removing the set of vertices $\{y_1,v_1,v_2,v_3,v_4\}$. If D is a ${\gamma }_t\left(T\right)$-set containing no leaf, then $y_3,y_2,v_2,v_3\in D$ and to dominate $v_5$, we may assume that $v_6\in D$. Thus, $D-\{v_2,v_3,y_1\}$ is a td-set of $T^{\prime }$, and thus ${\gamma }_t\left(T\right)\ge {\gamma }_t\left(T^{\prime }\right)+3$. On the other hand, any ${\gamma }_{dve}\left(T^{\prime }\right)$-set containing no leaf contains $y_3,v_6$, and thus can be extended to a dve-dominating set of T by adding $v_2,v_3,y_2.$ Therefore, ${\gamma }_{dve}\left(T\right)\le {\gamma }_{dve}\left(T^{\prime }\right)+3$. By Observation 3 (Item 2), we have ${\gamma }_t\left(T^{\prime }\right)={\gamma }_{dve}\left(T^{\prime }\right)$, and thus by our inductive hypothesis, $T^{\prime }\in \mathcal{T}$. As T can be obtained from $T^{\prime }$, by using Operation ${\mathcal{O}}_7,$, we deduce that $T\in \mathcal{T}$.

□