Extremal values on Zagreb indices of trees with given distance k-domination number

Abstract

Let $G=(V(G),E(G))$ be a graph. A set $D\subseteq V(G)$ is a distance k-dominating set of G if for every vertex $u\in V(G)\setminus D$, $d_G(u,v)\le k$ for some vertex $v\in D$, where k is a positive integer. The distance k-domination number ${\gamma }_k(G)$ of G is the minimum cardinality among all distance k-dominating sets of G. The first Zagreb index of G is defined as $M_1={\sum }_{u\in V(G)}{d}^2(u)$ and the second Zagreb index of G is $M_2={\sum }_{uv\in E(G)}d(u)d(v)$. In this paper, we obtain the upper bounds for the Zagreb indices of n-vertex trees with given distance k-domination number and characterize the extremal trees, which generalize the results of Borovićanin and Furtula (Appl. Math. Comput. 276:208–218, 2016). What is worth mentioning, for an n-vertex tree T, is that a sharp upper bound on the distance k-domination number ${\gamma }_k(T)$ is determined.

Introduction

Throughout this paper, all graphs considered are simple, undirected and connected. Let $G=(V,E)$ be a simple and connected graph, where $V=V(G)$ is the vertex set and $E=E(G)$ is the edge set of G. The eccentricity of v is defined as ${\epsilon }_G(v)=max\{d_G(u,v)\mid u\in V(G)\}$. The diameter of G is $diam(G)=max\{{\epsilon }_G(v)\mid v\in V(G)\}$. A path P is called a diameter path of G if the length of P is $diam(G)$. Denote by $N_G^i(v)$ the set of vertices with distance i from v in G, that is, $N_G^i(v)=\{u\in V(G)\mid d(u,v)=i\}$. In particular, $N_G^0(v)=\{v\}$ and $N_G^1(v)=N_G(v)$. A vertex $v\in V(G)$ is called a private k-neighbor of u with respect to D if ${\bigcup }_{i=0}^k{N}_G^i(v)\cap D=\{u\}$. That is, $d_G(v,u)\le k$ and $d_G(v,x)\ge k+1$ for any vertex $x\in D\setminus \{u\}$. The pendent vertex is the vertex of degree 1.

A chemical molecule can be viewed as a graph. In a molecular graph, the vertices represent the atoms of the molecule and the edges are chemical bonds. A topological index of a molecular graph is a mathematical parameter which is used for studying various properties of this molecule. The distance-based topological indices, such as the Wiener index [2, 3] and the Balaban index [4], have been extensively researched for many decades. Meanwhile the spectrum-based indices developed rapidly, such as the Estrada index [5], the Kirchhoff index [6] and matching energy [7]. The eccentricity-based topological indices, such as the eccentric distance sum [8], the connective eccentricity index [9] and the adjacent eccentric distance sum [10], were proposed and studied recently. The degree-based topological indices, such as the Randić index [11–13], the general sum-connectivity index [14, 15], the Zagreb indices [16], the multiplicative Zagreb indices [17, 18] and the augmented Zagreb index [19], where the Zagreb indices include the first Zagreb index $M_1={\sum }_{u\in V(G)}{d}^2(u)$ and the second Zagreb index $M_2={\sum }_{uv\in E(G)}d(u)d(v)$, represent one kind of the most famous topological indices. In this paper, we continue the work on Zagreb indices. Further study about the Zagreb indices can be found in [20–25]. Many researchers are interested in establishing the bounds for the Zagreb indices of graphs and characterizing the extremal graphs [1, 26–40].

A set $D\subseteq V(G)$ is a dominating set of G if, for any vertex $u\in V(G)\setminus D$, $N_G(u)\cap D\ne \mathrm{\varnothing }$. The domination number $\gamma (G)$ of G is the minimum cardinality of dominating sets of G. For $k\in N^{+}$, a set $D\subseteq V(G)$ is a distance k-dominating set of G if, for every vertex $u\in V(G)\setminus D$, $d_G(u,v)\le k$ for some vertex $v\in D$. The distance k-domination number ${\gamma }_k(G)$ of G is the minimum cardinality among all distance k-dominating sets of G [41, 42]. Every vertex in a minimum distance k-dominating set has a private k-neighbor. The domination number is the special case of the distance k-domination number for $k=1$. Two famous books [43, 44] written by Haynes et al. show us a comprehensive study of domination. The topological indices of graphs with given domination number or domination variations have attracted much attention of researchers [1, 45–47].

Borovićanin [1] showed the sharp upper bounds on the Zagreb indices of n-vertex trees with domination number γ and characterized the extremal trees. Motivated by [1], we describe the upper bounds for the Zagreb indices of n-vertex trees with given distance k-domination number and find the extremal trees. Furthermore, a sharp upper bound, in terms of $n,k$ and Δ, on the distance k-domination number ${\gamma }_k(T)$ for an n-vertex tree T is obtained in this paper.

Lemmas

In this section, we give some lemmas which are helpful to our results.

Lemma 2.1

([24, 48])

If T is an n-vertex tree, different from the star $S_n$, then $M_i(T)<M_i(S_n)$ for $i=1,2$.

In what follows, we present two graph transformations that increase the Zagreb indices.

Transformation I

([49])

Let T be an n-vertex tree ($n>3$) and $e=uv\in E(T)$ be a nonpendent edge. Assume that $T-uv=T_1\cup T_2$ with vertex $u\in V(T_1)$ and $v\in V(T_2)$. Let $T^{\prime }$ be the tree obtained by identifying the vertex u of $T_1$ with vertex v of $T_2$ and attaching a pendent vertex w to the u (=v) (see Figure 1). For the sake of convenience, we denote $T^{\prime }=\tau (T,uv)$.

Figure 1.

T and ${\mathit{T}}^{\mathbf{\prime }}$ in Transformation I .

Lemma 2.2

Let T be a tree of order n (≥3) and $T^{\prime }=\tau (T,uv)$. Then $M_i(T^{\prime })>M_i(T)$, $i=1,2$.

Proof

It is obvious that $d_{T^{\prime }}(u)=d_T(u)+d_T(v)-1$ and

$$\begin{array}{rl}{M}_1\left(T^{\prime }\right)-M_1(T)=& {(d_T(u)+d_T(v)-1)}^2+1-d_T^2(u)-d_T^2(v)\\ =& 2(d_T(u)-1)(d_T(v)-1)\\ >& 0.\end{array}$$

Let $x\in V(T)$ be a vertex different from u and v. Then

$$\begin{array}{rl}{M}_2\left(T^{\prime }\right)-M_2(T)=& (d_T(u)+d_T(v)-1)(\sum _{xu\in E(T_1)}{d}_T(x)+\sum _{xv\in E(T_2)}{d}_T(x)+1)\\ & -d_T(u)\sum _{xu\in E(T_1)}{d}_T(x)-d_T(v)\sum _{xv\in E(T_2)}{d}_T(x)-d_T(u)d_T(v)\\ =& (d_T(v)-1)\sum _{xu\in E(T_1)}{d}_T(x)+(d_T(u)-1)\sum _{xv\in E(T_2)}{d}_T(x)\\ & +d_T(u)+d_T(v)-1-d_T(u)d_T(v)\\ \ge & 2(d_T(v)-1)(d_T(u)-1)+d_T(u)+d_T(v)-1-d_T(u)d_T(v)\\ =& (d_T(v)-1)(d_T(u)-1)\\ >& 0.\end{array}$$

This completes the proof. □

Lemma 2.3

([50])

Let u and v be two distinct vertices in G. $u_1,u_2,\dots ,u_r$ are the pendent vertices adjacent to u and $v_1,v_2,\dots ,v_t$ are the pendent vertices adjacent to v. Define $G^{\prime }=G-\{v{v}_1,v{v}_2,\dots ,v{v}_t\}+\{u{v}_1,u{v}_2,\dots ,u{v}_t\}$ and $G^{″}=G-\{u{u}_1,u{u}_2,\dots ,u{u}_r\}+\{v{u}_1,v{u}_2,\dots ,v{u}_r\}$, as shown in Figure 2. Then either $M_i(G^{\prime })>M_i(G)$ or $M_i(G^{″})>M_i(G)$, $i=1,2$.

Figure 2.

G , ${\mathit{G}}^{\mathbf{\prime }}$ and ${\mathit{G}}^{\mathbf{″}}$ in Lemma  2.3 .

Lemma 2.4

([51])

For a connected graph G of order n with $n\ge k+1$, ${\gamma }_k(G)\le \lfloor \frac{n}{k+1}\rfloor$.

Let G be a connected graph of order n. If ${\gamma }_k(G)\ge 2$, then $n\ge k+1$. Otherwise, ${\gamma }_k(G)=1$, a contradiction. Hence, by Lemma 2.4, we have ${\gamma }_k(G)\le \lfloor \frac{n}{k+1}\rfloor$ and $n\ge (k+1){\gamma }_k$ for any connected graph G of order n if ${\gamma }_k(G)\ge 2$.

Lemma 2.5

Let T be an n-vertex tree with distance k-domination number ${\gamma }_k\ge 2$. Then $\mathrm{△}\le n-k{\gamma }_k$.

Proof

Suppose that $\mathrm{△}\ge n-k{\gamma }_k+1$. Let $v\in V(T)$ be the vertex such that $d(v)=\mathrm{△}$ and $N(v)=\{v_1,\dots ,v_{\mathrm{△}}\}$. Denote by $T^i$ the component of $T-v$ containing the vertex $v_i$, $i=1,\dots ,\mathrm{△}$. Let D be a minimum distance k-dominating set of T,

$$S_1=\{i\mid i\in \{1,2,\dots ,\mathrm{△}\},0\le {\epsilon }_{T^i}(v_i)\le k-1\}$$

and

$$S_2=\{i\mid i\in \{1,2,\dots ,\mathrm{△}\},{\epsilon }_{T^i}(v_i)\ge k\}.$$

Clearly, $|S_2|\ge 1$. If not, $\{v\}$ is a distance k-dominating set of T, which contradicts ${\gamma }_k\ge 2$. If $|S_1|=0$, then ${\epsilon }_{T^i}(v_i)\ge k$ for $i=1,\dots ,\mathrm{△}$, so $|V(T^i)\cap D|\ge 1$. Therefore, ${\gamma }_k\ge \mathrm{△}\ge n-k{\gamma }_k+1$, which implies that ${\gamma }_k\ge \frac{n+1}{k+1}$. Since ${\gamma }_k\ge 2$, ${\gamma }_k\le \lfloor \frac{n}{k+1}\rfloor$ by Lemma 2.4, a contradiction. Thus, $|S_1|\ge 1$. Let $i_1\in S_1$ and

$${\epsilon }_{T^{i_1}}(v_{i_1})=max\{{\epsilon }_{T^i}(v_i)\mid i\in S_1\}=\lambda .$$

Then $0\le \lambda \le k-1$, so $|S_2|\le \lfloor \frac{n-\mathrm{△}-1-\lambda }{k}\rfloor \le \lfloor \frac{k{\gamma }_k-2}{k}\rfloor \le {\gamma }_k-1$.

If $V(T^i)\cap D=D_1\ne \mathrm{\varnothing }$ for some $i\in S_1$, then $D-D_1+\{v\}$ is a distance k-dominating set according to the definition of $S_1$. Thus, we assume that $V(T^i)\cap D=\mathrm{\varnothing }$ for each $i\in S_1$. Similarly, suppose that $D^{\prime }\cap V(T^{i_1})=\mathrm{\varnothing }$ where $D^{\prime }$ is a minimum distance k-dominating set of the tree $T^{\prime }=T-{\bigcup }_{i\in S_1\setminus \{i_1\}}V(T^i)$.

We claim that $D^{\prime }$ is a distance k-dominating set of T. Let $y\in V(T^{i_1})$ be the vertex such that $d(v_{i_1},y)=\lambda$ and $y^{\prime }\in D_1^{\prime }={\bigcup }_{i=0}^k{N}_{T^{\prime }}^i(y)\cap D^{\prime }$. Then $y^{\prime }\in V(T^{\prime })\setminus V(T^{i_1})$ and $d(y,y^{\prime })=d(y,v)+d(v,y^{\prime })\le k$, so, for $x\in {\bigcup }_{i\in S_1\setminus \{i_1\}}V(T^i)$, we have $d(x,y^{\prime })=d(x,v)+d(v,y^{\prime })\le d(y,v)+d(v,y^{\prime })\le k$. Hence, all the vertices in ${\bigcup }_{i\in S_1\setminus \{i_1\}}V(T^i)$ can be dominated by $y^{\prime }\in D^{\prime }$. Therefore, $D^{\prime }$ is a distance k-dominating set of T, so the claim is true.

In view of

$$k+1<(k+1)|S_2|+\lambda +2\le \left|V\left(T^{\prime }\right)\right|\le n-|S_1|+1=n-\mathrm{△}+|S_2|+1,$$

one has

$$\begin{array}{rl}{\gamma }_k& \le \left|D^{\prime }\right|\\ & \le \lfloor \frac{n-\mathrm{△}+|S_2|+1}{k+1}\rfloor (\text{by Lemma 2.4})\\ & \le \lfloor \frac{(k+1){\gamma }_k-1}{k+1}\rfloor (\text{since }\mathrm{△}\ge n-k{\gamma }_k+1,|S_2|\le {\gamma }_k-1)\\ & <{\gamma }_k,\end{array}$$

a contradiction as desired. □

Determining the bound on the distance k-domination number of a connected graph is an attractive problem. In Lemma 2.5, an upper bound for the distance k-domination number of a tree is characterized. Namely, if T is an n-vertex tree with distance k-domination number ${\gamma }_k\ge 2$, then ${\gamma }_k(T)\le \frac{n-\mathrm{\Delta }(T)}{k}$.

Let ${\mathcal{T}}_{n,k,{\gamma }_k}$ be the set of all n-vertex trees with distance k-domination number ${\gamma }_k$ and $S_{n-k{\gamma }_k+1}$ be the star of order $n-k{\gamma }_k+1$ with pendent vertices $v_1,v_2,\dots ,v_{n-k{\gamma }_k}$. Denote by $T_{n,k,{\gamma }_k}$ the tree formed from $S_{n-k{\gamma }_k}$ by attaching a path $P_{k-1}$ to $v_1$ and attaching a path $P_k$ to $v_i$ for each $i\in \{2,\dots ,{\gamma }_k\}$, as shown in Figure 3. Then $T_{n,k,{\gamma }_k}\in {\mathcal{T}}_{n,k,{\gamma }_k}$. Even more noteworthy is the notion that ${\gamma }_k(T_{n,k,{\gamma }_k})={\gamma }_k=\frac{n-\mathrm{\Delta }(T_{n,k,{\gamma }_k})}{k}$. It implies that the upper bound on the distance k-domination number mentioned in the above paragraph is sharp.

Figure 3.

${\mathit{T}}_{\mathit{n}\mathbf{,}\mathit{k}\mathbf{,}{\mathit{\gamma }}_{\mathit{k}}}$ .

The Zagreb indices of $T_{n,k,{\gamma }_k}$ are computed as

$$M_1(T_{n,k,{\gamma }_k})=(n-k{\gamma }_k)(n-k{\gamma }_k+1)+4(k{\gamma }_k-1)$$

and

$$M_2(T_{n,k,{\gamma }_k})=\{\begin{array}{cc}(n-k{\gamma }_k)[n-(k-1){\gamma }_k]+(4k-2){\gamma }_k-4\hfill & \text{if }k\ge 2,\hfill \\ 2(n-\gamma +1)(\gamma -1)+(n-\gamma )(n-2\gamma +1)\hfill & \text{if }k=1.\hfill \end{array}$$

For $k=1$, the distance k-domination number ${\gamma }_1(G)$ is the domination number $\gamma (G)$. Furthermore, the upper bounds on the Zagreb indices of an n-vertex tree with domination number were studied in [1], so we only consider $k\ge 2$ in the following.

Lemma 2.6

([52])

T be a tree on $(k+1)n$ vertices. Then ${\gamma }_k(T)=n$ if and only if at least one of the following conditions holds:

1. T is any tree on $k+1$ vertices;

2. $T=R\circ k$ for some tree R on $n\ge 1$ vertices, where $R\circ k$ is the graph obtained by taking one copy of R and $|V(R)|$ copies of the path $P_{k-1}$ of length $k-1$ and then joining the ith vertex of R to exactly one end vertex in the ith copy of $P_{k-1}$.

Lemma 2.7

Let T be an n-vertex tree with distance k-domination number ${\gamma }_k(T)\ge 3$. If $n=(k+1){\gamma }_k$, then

$$M_1(T)\le {\gamma }_k({\gamma }_k+1)+4(k{\gamma }_k-1)$$

and

$$M_2(T)\le 2{\gamma }_k^2+(4k-2){\gamma }_k-4,$$

with equality if and only if $T\cong T_{n,k,{\gamma }_k}$.

Proof

When $n=(k+1){\gamma }_k$, $T=R\circ k$ for some tree R on ${\gamma }_k$ vertices by Lemma 2.6. Assume that $V(R)=\{v_1,\dots ,v_{{\gamma }_k}\}$. Then $d_R(v_i)=d_T(v_i)-1$. It is well known that ${\sum }_{i=1}^{n}d(u_i)=2(n-1)$ for any n-vertex tree with vertex set $\{u_1,\dots ,u_n\}$. Hence, ${\sum }_{i=1}^{{\gamma }_k}{d}_R(v_i)=2({\gamma }_k-1)$. By the definition of the first Zagreb index, we have

$$\begin{array}{rcl}{M}_1(T)& =& \sum _{i=1}^{{\gamma }_k}{d}_T^2(v_i)+\sum _{x\in V(T)\setminus V(R)}{d}_T^2(x)\\ & =& \sum _{i=1}^{{\gamma }_k}{(d_T(v_i)-1)}^2+\sum _{x\in V(T)\setminus V(R)}{d}_T^2(x)+2\sum _{i=1}^{{\gamma }_k}(d_T(v_i)-1)+{\gamma }_k\\ & =& M_1(R)+4(k-1){\gamma }_k+{\gamma }_k+2\sum _{i=1}^{{\gamma }_k}{d}_R(v_i)+{\gamma }_k\\ & \le & M_1(S_{{\gamma }_k})+4(k-1){\gamma }_k+2{\gamma }_k+4({\gamma }_k-1)\\ & =& {\gamma }_k({\gamma }_k+1)+4(k{\gamma }_k-1).\end{array}$$

The equality holds if and only if $R\cong S_{{\gamma }_k}$, that is, $T\cong T_{n,k,{\gamma }_k}$. We have

$$\begin{array}{rl}{M}_2(T)=& \sum _{xy\in E(R)}{d}_T(x)d_T(y)+\sum _{xy\in E(T)\setminus E(R)}{d}_T(x)d_T(y)\\ =& \sum _{xy\in E(R)}(d_T(x)-1)(d_T(y)-1)+\sum _{xy\in E(R)}(d_T(x)+d_T(y)-1)\\ & +\sum _{xy\in E(T)\setminus E(R)}{d}_T(x)d_T(y)\\ =& M_2(R)+\sum _{x\in V(R)}{d}_T(x)(d_T(x)-1)-({\gamma }_k-1)\\ & +\sum _{x\in V(R)}2d_T(x)+4(k-2){\gamma }_k+2{\gamma }_k\\ =& M_2(R)+\sum _{x\in V(R)}{(d_T(x)-1)}^2+3\sum _{x\in V(R)}(d_T(x)-1)+4k{\gamma }_k-5{\gamma }_k-1\\ =& M_2(R)+M_1(R)+6({\gamma }_k-1)+4k{\gamma }_k-5{\gamma }_k+1\\ \le & M_2(S_{{\gamma }_k})+M_1(S_{{\gamma }_k})+4k{\gamma }_k+{\gamma }_k-5\\ =& 2{\gamma }_k^2+(4k-2){\gamma }_k-4.\end{array}$$

The equality holds if and only if $R\cong S_{{\gamma }_k}$. As a consequence, $T\cong T_{n,k,{\gamma }_k}$. □

Lemma 2.8

Let G be a graph which has a maximum value of the Zagreb indices among all n-vertex connected graphs with distance k-domination number and $S_G=\{v\in V(G)\mid d_G(v)=1,{\gamma }_k(G-v)={\gamma }_k(G)\}$. If $S_G\ne \mathrm{\varnothing }$, then $|N_G(S_G)|=1$.

Proof

Suppose that $|N_G(S_G)|\ge 2$ and u and v are two distinct vertices in $N_G(S_G)$. $x_1,x_2,\dots ,x_r$ are the pendent vertices adjacent to u and $y_1,y_2,\dots ,y_t$ are the pendent vertices adjacent to v, where $r\ge 1$ and $t\ge 1$. Let D be a minimum distance k-dominating set of G. If $x_i\in D$ for some $i\in \{1,\dots ,r\}$, then $D-x_i+u$ is a distance k-dominating set of T. Hence, we assume that $x_i\notin D$, $i=1,\dots ,r$. Similarly, $y_i\notin D$ for $1\le i\le t$. Define $G_1=G-\{v{y}_1\}+\{u{y}_1\}$ and $G_2=G-\{u{x}_1\}+\{v{x}_1\}$. Then ${\gamma }_k(G_1)={\gamma }_k(G_2)={\gamma }_k(G)$. In addition, we have either $M_i(G_1)>M_i(G)$ or $M_i(G_2)>M_i(G)$, $i=1,2$, by a similar proof of Lemma 2.3 and thus omitted here (for reference, see the Appendix). It follows a contradiction, as desired. □

Main results

In this section, we give upper bounds on the Zagreb indices of a tree with given order n and distance k-domination number ${\gamma }_k$. If $P=v_0{v}_1\cdots v_d$ is a diameter path of an n-vertex tree T, then denote by $T_i$ the component of $T-\{v_{i-1}{v}_i,v_i{v}_{i+1}\}$ containing $v_i$, $i=1,2,\dots ,d-1$. By Lemma 2.1, we obtain Theorem 3.1 directly.

Theorem 3.1

Let T be an n-vertex tree and ${\gamma }_k(T)=1$. Then $M_1(T)\le n(n-1)$ and $M_2(T)\le {(n-1)}^2$. The equality holds if and only if $T\cong S_n$.

Let $T_{n,k,2}^i$ be the tree obtained from the path $P_{2k+2}=v_0\cdots v_{2k+1}$ by joining $n-2(k+1)$ pendent vertices to $v_i$, where $i\in \{1,\dots ,2k\}$.

Theorem 3.2

If T is an n-vertex tree with distance k-domination number ${\gamma }_k(T)=2$, then

$$M_1(T)\le (n-2k)(n-2k+1)+4(2k-1),$$

with equality if and only if $T\cong T_{n,k,2}^i$, where $i\in \{1,\dots ,k\}$. Also,

$$M_2(T)\le (n-2k)(n-2k+2)+8k-8,$$

with equality if and only if $T\cong T_{n,k,2}^i$, where $i\in \{2,\dots ,k\}$.

Proof

Assume that $T\in {\mathcal{T}}_{n,k,2}$ is the tree that maximizes the Zagreb indices and $P=v_0{v}_1\cdots v_d$ is a diameter path of T. If $d\le 2k$, then $\{v_{\lfloor \frac{d}{2}\rfloor }\}$ is a distance k-dominating set of T, a contradiction to ${\gamma }_k(T)=2$. If $d\ge 2k+2$, define $T^{\prime }=\tau (T,v_i{v}_{i+1})$, where $i\in \{1,\dots ,d-2\}$. Then $T^{\prime }\in {\mathcal{T}}_{n,k,2}$. By Lemma 2.2, we have $M_i(T^{\prime })>M_i(T)$, $i=1,2$, a contradiction. Hence, $d=2k+1$.

If $T_i$ is not a star for some $i\in \{1,2,\dots ,d-1\}$, then there exists an n-vertex tree $T^{\prime }$ in ${\mathcal{T}}_{n,k,2}$ such that $M_i(T^{\prime })>M_i(T)$ for $i=1,2$ by Lemma 2.2, a contradiction. Besides, $T\cong T_{n,k,2}^i$ for some $i\in \{1,\dots ,d-1\}$ by Lemma 2.3.

Since $M_1(T_{n,k,2}^i)=M_1(T_{n,k,2}^j)$ for $1\le i\ne j\le d-1$ and $T_{n,k,2}^i\cong T_{n,k,2}^{d-i}$ for $k+1\le i\le d-1$, we get $T\cong T_{n,k,2}^i$, $i\in \{1,\dots ,k\}$. By direct computation, one has $M_1(T)=M_1(T_{n,k,2}^i)=(n-2k)(n-2k+1)+4(2k-1)$, $i\in \{1,\dots ,k\}$. In addition, $M_2(T_{n,k,2}^1)=M_2(T_{n,k,2}^{d-1})<M_2(T_{n,k,2}^2)=\cdots =M_2(T_{n,k,2}^{d-2})$ and $T_{n,k,2}^i\cong T_{n,k,2}^{d-i}$ for $i\in \{k+1,\dots ,d-2\}$. Hence, $T\cong T_{n,k,2}^i$, where $i\in \{2,\dots ,k\}$. Moreover, $M_2(T)=M_2(T_{n,k,2}^i)=(n-2k)(n-2k+2)+8k-8$. This completes the proof. □

Lemma 3.3

Let tree $T\in {\mathcal{T}}_{n,k,3}$. Then

$$M_1(T)\le (n-3k)(n-3k+1)+4(3k-1)$$

and

$$M_2(T)\le (n-3k)(n-3k+3)+12k-10,$$

with equality if and only if $T\cong T_{n,k,3}$.

Proof

Assume that $T\in {\mathcal{T}}_{n,k,3}$. We complete the proof by induction on n. By Lemma 2.4, we have $n\ge (k+1){\gamma }_k$. This lemma is true for $n=(k+1){\gamma }_k$ by Lemma 2.7. Suppose that $n>3(k+1)$ and the statement holds for $n-1$ in the following.

Let D be a minimum distance k-dominating set of T and $P=v_0{v}_1\cdots v_d$ be a diameter path of T. Then $d\ge 2k+2$. Otherwise, $\{v_k,v_{k+1}\}$ is a distance k-dominating set, a contradiction. Note that ${\bigcup }_{i=0}^k{N}_T^i(v_0)\cap D\ne \mathrm{\varnothing }$ and ${\bigcup }_{i=0}^k{N}_T^i(v_0)\subseteq ({\bigcup }_{i=0}^{k-1}V(T_i)\cup \{v_k\})$. Hence, $({\bigcup }_{i=0}^{k-1}V(T_i)\cup \{v_k\})\cap D\ne \mathrm{\varnothing }$. However, ${\bigcup }_{i=0}^k{N}_T^i(x)\subseteq {\bigcup }_{i=0}^k{N}_T^i(v_k)$ for $x\in {\bigcup }_{i=0}^{k}V(T_i)\setminus \{v_k\}$, so we assume that $v_k\in D$ and $({\bigcup }_{i=0}^{k}V(T_i)\setminus \{v_k\})\cap D=\mathrm{\varnothing }$. Similarly, $v_{d-k}\in D$ and $({\bigcup }_{i=d-k}^{d}V(T_i)\setminus \{v_{d-k}\})\cap D=\mathrm{\varnothing }$. Suppose that $v_0=u_1$, $v_d=u_2,\dots ,u_m$ are the pendent vertices of T and $S_T=\{u_i\mid 1\le i\le m,{\gamma }_k(T-u_i)={\gamma }_k(T)\}$. We have the following claim.

Claim 1

$S_T\ne \mathrm{\varnothing }$.

Proof

Assume that $S_T=\mathrm{\varnothing }$. Namely, ${\gamma }_k(T-u_i)={\gamma }_k(T)-1$ for each $i\in \{1,\dots ,m\}$. If $D\setminus \{w_i\}$ is a minimum distance k-dominating set of the tree $T-u_i$, where $w_i\in D$, then $w_i\ne w_j$ for $1\le i\ne j\le m$. Otherwise, ${\gamma }_k(T-u_i)={\gamma }_k(T)$ or ${\gamma }_k(T-u_j)={\gamma }_k(T)$, a contradiction. It follows that $m\le {\gamma }_k$.

If $d_T(v_i)\ge 3$ for some $i\in \{2,\dots ,k,d-k,\dots ,d-1\}$, then $V(T_i)\cap \{u_3,\dots ,u_m\}\ne \mathrm{\varnothing }$. In view of $\{v_k,v_{d-k}\}\subseteq D$, we have ${\gamma }_k(T-x)={\gamma }_k(T)$ for $x\in V(T_i)\cap \{u_3,\dots ,u_m\}$, a contradiction. Hence, $d_T(v_i)=2$ for $i\in \{2,\dots ,k,d-k,\dots ,d-1\}$.

Since ${\gamma }_k(T-v_0)={\gamma }_k(T)-1$, $v_1$ must be dominated by the vertices in $D\setminus \{v_k\}$. Bearing in mind that $({\bigcup }_{i=0}^{k}V(T_i)\setminus \{v_k\})\cap D=\mathrm{\varnothing }$, one has $v_{k+1}\in D$. The same applies to $v_{d-k-1}$. Hence, $\{v_k,v_{k+1},v_{d-k-1},v_{d-k}\}\subseteq D$. If $d>2k+2$, then the vertices $v_k$, $v_{k+1}$, $v_{d-k-1}$ and $v_{d-k}$ are different from each other, a contradiction to ${\gamma }_k(T)=3$. As a consequence, $d=2k+2$ and thus $D=\{v_k,v_{k+1},v_{d-k}\}$.

If $d_T(v_{k+1})=2$, then $T\cong P_{2k+3}$ and $\{v_k,v_{d-k}\}$ is a distance k-dominating set, a contradiction. It follows that $d_T(v_{k+1})\ge 3$. Hence, $m\ge 3={\gamma }_k$. Recalling that $m\le {\gamma }_k=3$, we have $m=3$, which implies that $T_{k+1}$ is a path with end vertices $v_{k+1}$ and $u_3$. If $d(v_{k+1},u_3)>k$, then $u_3$ cannot be dominated by the vertices in D. If $d(v_{k+1},u_3)<k$, then $D\setminus \{v_{k+1}\}$ is a distance k-dominating set, a contradiction. Therefore, $d(v_{k+1},u_3)=k$. We conclude that $|V(T)|=3(k+1)$, which contradicts $n>3(k+1)$, so Claim 1 is true. □

Considering $S_T\ne \mathrm{\varnothing }$ for $T\in {\mathcal{T}}_{n,k,3}$, the tree among ${\mathcal{T}}_{n,k,3}$ that maximizes the Zagreb indices must be in the set $\{T^{\ast }\in {\mathcal{T}}_{n,k,3}\mid |N_{T^{\ast }}(S_{T^{\ast }})|=1\}$ by Lemma 2.8. To determine the extremal trees among ${\mathcal{T}}_{n,k,3}$, we assume that $T\in \{T^{\ast }\in {\mathcal{T}}_{n,k,3}\mid |N_{T^{\ast }}(S_{T^{\ast }})|=1\}$ in what follows.

Let $u_i$ be a pendent vertex such that ${\gamma }_k(T-u_i)={\gamma }_k(T)$ and s be the unique vertex adjacent to $u_i$. By Lemma 2.5, $d_T(s)\le \mathrm{△}\le n-k{\gamma }_k$. Define $A=\{x\in V(T)\mid d_T(x)=1,xs\notin E(T)\}$ and $B=\{x\in V(T)\mid d_T(x)\ge 2,xs\notin E(T)\}$. Then ${\gamma }_k(T-x)={\gamma }_k(T)-1$ for all $x\in A$. As a consequence, $|A|\le {\gamma }_k$ from the proof of Claim 1. By the induction hypothesis,

$$\begin{array}{rcl}{M}_1(T)& =& M_1(T-u_i)+2d(s)\\ & \le & (n-1-k{\gamma }_k)(n-1-k{\gamma }_k+1)+4(k{\gamma }_k-1)+2(n-k{\gamma }_k)\\ & =& (n-k{\gamma }_k)(n-k{\gamma }_k+1)+4(k{\gamma }_k-1).\end{array}$$

The equality holds if and only if $T-u_i\cong T_{n-1,k,{\gamma }_k}$ and $d_T(s)=\mathrm{△}=n-k{\gamma }_k$, i.e., $T\cong T_{n,k,{\gamma }_k}$.

Note that $|A|+|B|=n-1-d_T(s)$ and $|A|\le {\gamma }_k$. Therefore, $|B|=n-1-d_T(s)-|A|\ge n-1-d_T(s)-{\gamma }_k$ and

$$\sum _{xs\notin E(T)}d(x)\ge |A|+2|B|=(|A|+|B|)+|B|\ge 2(n-1-d_T(s))-{\gamma }_k.$$

By the above inequality and the definition of $M_2$, we have

$$\begin{array}{rcl}{M}_2(T)& =& M_2(T-u_i)+\sum _{v\in V(T)}{d}_T(v)-\sum _{xs\notin E(T)}{d}_T(x)-1\\ & \le & M_2(T-u_i)+2(n-1)-2(n-1-d_T(s))+{\gamma }_k-1\end{array}$$ 1

$$\begin{array}{rcl}& \le & (n-1-k{\gamma }_k)[n-1-(k-1){\gamma }_k]+(4k-2){\gamma }_k-4\\ & & +2(n-k{\gamma }_k)+{\gamma }_k-1(\text{since }{d}_T(s)\le \mathrm{△}\le n-k{\gamma }_k)\\ & =& (n-k{\gamma }_k)[n-(k-1){\gamma }_k]+(4k-2){\gamma }_k-4.\end{array}$$ 2

The equality (1) holds if and only if $|A|={\gamma }_k$, $|B|=n-1-d_T(s)-{\gamma }_k$ and $d_T(x)=2$ for $x\in B$. The equality (2) holds if and only if $T-u_i\cong T_{n-1,k,{\gamma }_k}$ and $d_T(s)=\mathrm{△}=n-k{\gamma }_k$. Hence, $M_2(T)\le (n-k{\gamma }_k)[n-(k-1){\gamma }_k]+(4k-2){\gamma }_k-4$ with equality if and only if $T\cong T_{n,k,{\gamma }_k}$. □

Theorem 3.4

Let T be a tree of order n with distance k-domination number ${\gamma }_k$ (≥3). Then

$$M_1(T)\le (n-k{\gamma }_k)(n-k{\gamma }_k+1)+4(k{\gamma }_k-1)$$

and

$$M_2(T)\le (n-k{\gamma }_k)[n-(k-1){\gamma }_k]+(4k-2){\gamma }_k-4,$$

with equality if and only if $T\cong T_{n,k,{\gamma }_k}$.

Proof

Let $T\in {\mathcal{T}}_{n,k,{\gamma }_k}$ and $P=v_0{v}_1\cdots v_d$ be a diameter path of T. Define $S_T=\{u\in V(T)\mid d_T(u)=1,{\gamma }_k(T-u)={\gamma }_k(T)\}$. If $S_T=\mathrm{\varnothing }$, then ${\gamma }_k(T-v_i)={\gamma }_k(T)-1$ for $i=0,d$. If $S_T\ne \mathrm{\varnothing }$, then we suppose that $T\in \{T^{\ast }\in {\mathcal{T}}_{n,k,{\gamma }_k}\mid |N_{T^{\ast }}(S_{T^{\ast }})|=1\}$ by Lemma 2.8 for establishing the maximum Zagreb indices of trees among ${\mathcal{T}}_{n,k,{\gamma }_k}$. If $v_d\in S_T\ne \mathrm{\varnothing }$, then ${\gamma }_k(T-v_0)={\gamma }_k(T)-1$, which implies that ${\gamma }_k(T-v_0)={\gamma }_k(T)-1$ or ${\gamma }_k(T-v_d)={\gamma }_k(T)-1$. Assume that ${\gamma }_k(T-v_0)={\gamma }_k(T)-1$. Then there is a minimum distance k-dominating set D of T such that $\{v_k,v_{k+1},v_{d-k}\}\subseteq D$ from the proof of Lemma 3.3.

Let $T^{\prime }$ be the tree obtained from T by applying Transformation I on $T_i$ repeatedly for $i=1,\dots ,k$ such that $T_i^{\prime }\cong S_{|V(T_i^{\prime })|}$, where $T_i^{\prime }$ is the component of $T^{\prime }-\{v_{i-1}{v}_i,v_i{v}_{i+1}\}$ containing $v_i$, $i=1,\dots ,k$ (see Figure 4). Then $T^{\prime }\in {\mathcal{T}}_{n,k,{\gamma }_k}$. By Lemma 2.2, we have $M_i(T)\le M_i(T^{\prime })$, $i=1,2$, with equality if and only if $T\cong T^{\prime }$.

Figure 4.

T , ${\mathit{T}}^{\mathbf{\prime }}$ , ${\mathit{T}}^{\mathbf{″}}$ and ${\mathit{T}}^{\mathbf{‴}}$ .

By Lemma 2.3, for some $i_0,i_1\in \{1,\dots ,k\}$, define

$$\begin{array}{rl}{T}^{″}=& T^{\prime }-\bigcup _{i\in \{1,\dots ,k\}\setminus \{i_0\}}\{v_{i}x\mid x\in N_{T^{\prime }}(v_i)\setminus \{v_{i-1},v_{i+1}\}\}\\ & +\bigcup _{i\in \{1,\dots ,k\}\setminus \{i_0\}}\{v_{i_0}x\mid x\in N_{T^{\prime }}(v_i)\setminus \{v_{i-1},v_{i+1}\}\}\end{array}$$

and

$$\begin{array}{rl}{\tilde{T}}^{″}=& T^{\prime }-\bigcup _{i\in \{1,\dots ,k\}\setminus \{i_1\}}\{v_{i}x\mid x\in N_{T^{\prime }}(v_i)\setminus \{v_{i-1},v_{i+1}\}\}\\ & +\bigcup _{i\in \{1,\dots ,k\}\setminus \{i_1\}}\{v_{i_1}x\mid x\in N_{T^{\prime }}(v_i)\setminus \{v_{i-1},v_{i+1}\}\}.\end{array}$$

Then one has $M_1(T^{\prime })\le M_1(T^{″})$ with equality if and only if $T^{\prime }\cong T^{″}$ and $M_2(T^{\prime })\le M_2({\tilde{T}}^{″})$ with equality if and only if $T^{\prime }\cong {\tilde{T}}^{″}$.

Suppose that $|N_{T^{″}}(v_{i_0})\setminus \{v_{i_0-1},v_{i_0+1}\}|=|N_{{\tilde{T}}^{″}}\setminus \{v_{i_1-1},v_{i_1+1}\}|=m$, $m\ge 0$. Let

$$\begin{array}{rl}{T}^{‴}=& T^{″}-\{v_{i_0}x\mid x\in N_{T^{″}}(v_{i_0})\setminus \{v_{i_0-1},v_{i_0+1}\}\}\\ & +\{v_{k+1}x\mid x\in N_{T^{″}}(v_{i_0})\setminus \{v_{i_0-1},v_{i_0+1}\}\}\\ =& {\tilde{T}}^{″}-\{v_{i_1}x\mid x\in N_{{\tilde{T}}^{″}}(v_{i_1})\setminus \{v_{i_1-1},v_{i_1+1}\}\}\\ & +\{v_{k+1}x\mid x\in N_{{\tilde{T}}^{″}}(v_{i_1})\setminus \{v_{i_1-1},v_{i_1+1}\}\}.\end{array}$$

Then D is a minimum distance k-dominating set of $T^{‴}$ and $d_{T^{‴}}(v_i)=2$ for $i=1,\dots ,k$. Assume that ${\mathit{PN}}_{k,D}(x)$ is the set of all private k-neighbors of x with respect to D in $T^{‴}$. It is clear that the vertices in ${\bigcup }_{i=0}^k{N}_{T^{‴}}^i(v_k)\setminus \{v_0,\dots ,v_k\}$ can be dominated by $v_{k+1}\in D$. Thus, $D\setminus \{v_k\}$ is a distance k-dominating set of tree $T^{‴}-\{v_0,\dots ,v_k\}$. In addition, ${\mathit{PN}}_{k,D}(v_{k+1})\subseteq V(T^{‴})\setminus \{v_0,\dots ,v_k\}$, which means that $D\setminus \{v_k\}$ is a minimum distance k-dominating set of $T^{‴}-\{v_0,\dots ,v_k\}$. So ${\gamma }_k(T^{‴}-\{v_0,\dots ,v_k\})={\gamma }_k-1$. Analogously, ${\gamma }_k(T^{‴}-\{v_0,\dots ,v_{k-1}\})={\gamma }_k-1$.

By the definition of the first Zagreb index, we get

$$\begin{array}{rl}{M}_1\left(T^{‴}\right)-M_1\left(T^{″}\right)& =4+{(d_{T^{″}}(v_{k+1})+m)}^2-{(2+m)}^2-d_{T^{″}}^2(v_{k+1})\\ & =2m(d_{T^{″}}(v_{k+1})-2)\\ & \ge 0,\end{array}$$

so $M_1(T^{‴})-M_1(T^{″})=0$ if and only if at least one of the following conditions holds:

1. $m=0$, which implies that $T^{″}\cong T^{‴}$;

2. $d_{T^{″}}(v_{k+1})=2$.

If $i_1=1$, then

$$\begin{array}{rl}{M}_2\left(T^{‴}\right)-M_2\left({\tilde{T}}^{″}\right)=& 6+(d_{{\tilde{T}}^{″}}(v_{k+1})+m)(m+\sum _{x\in N_{{\tilde{T}}^{″}}(v_{k+1})}{d}_{{\tilde{T}}^{″}}(x))\\ & -(m+2)(m+3)-d_{{\tilde{T}}^{″}}(v_{k+1})\sum _{x\in N_{{\tilde{T}}^{″}}(v_{k+1})}{d}_{{\tilde{T}}^{″}}(x)\\ =& m[d_{{\tilde{T}}^{″}}(v_{k+1})+\sum _{x\in N_{{\tilde{T}}^{″}}(v_{k+1})}{d}_{{\tilde{T}}^{″}}(x)-5]\\ \ge & 0,\end{array}$$

with equality if and only if $m=0$, that is, ${\tilde{T}}^{″}\cong T^{‴}$. If $i_1\ne 1$ and $i_1\ne k$, then

$$\begin{array}{rl}{M}_2\left(T^{‴}\right)-M_2\left({\tilde{T}}^{″}\right)=& 8+(d_{{\tilde{T}}^{″}}(v_{k+1})+m)(m+\sum _{x\in N_{{\tilde{T}}^{″}}(v_{k+1})}{d}_{{\tilde{T}}^{″}}(x))\\ & -(m+2)(m+4)-d_{{\tilde{T}}^{″}}(v_{k+1})\sum _{x\in N_{{\tilde{T}}^{″}}(v_{k+1})}{d}_{{\tilde{T}}^{″}}(x)\\ =& m[d_{{\tilde{T}}^{″}}(v_{k+1})+\sum _{x\in N_{{\tilde{T}}^{″}}(v_{k+1})}{d}_{{\tilde{T}}^{″}}(x)-6]\\ \ge & 0.\end{array}$$

Also, $M_2(T^{‴})-M_2({\tilde{T}}^{″})=0$ if and only if at least one of the following conditions holds:

1. $m=0$, namely, ${\tilde{T}}^{″}\cong T^{‴}$;

2. $d_{{\tilde{T}}^{″}}(v_k)=d_{{\tilde{T}}^{″}}(v_{k+1})=d_{{\tilde{T}}^{″}}(v_{k+2})=2$.

If $i_1\ne 1$ and $i_1=k$, then

$$\begin{array}{rl}{M}_2\left(T^{‴}\right)-M_2\left({\tilde{T}}^{″}\right)=& 4+(d_{{\tilde{T}}^{″}}(v_{k+1})+m)(m+2+\sum _{x\in N_{{\tilde{T}}^{″}}(v_{k+1})\setminus \{v_k\}}{d}_{{\tilde{T}}^{″}}(x))\\ & -(m+2)(m+2)-d_{{\tilde{T}}^{″}}(v_{k+1})(\sum _{x\in N_{{\tilde{T}}^{″}}(v_{k+1})\setminus \{v_k\}}{d}_{{\tilde{T}}^{″}}(x)+m+2)\\ =& m(\sum _{x\in N_{{\tilde{T}}^{″}}(v_{k+1})\setminus \{v_k\}}{d}_{{\tilde{T}}^{″}}(x)-2)\\ \ge & 0.\end{array}$$

As a result, $M_2(T^{‴})-M_2({\tilde{T}}^{″})=0$ if and only if at least one of the following conditions holds:

1. $m=0$, which implies that ${\tilde{T}}^{″}\cong T^{‴}$;

2. $d_{{\tilde{T}}^{″}}(v_{k+1})=d_{{\tilde{T}}^{″}}(v_{k+2})=2$.

In what follows, we prove $M_1(T^{‴})\le (n-k{\gamma }_k)(n-k{\gamma }_k+1)+4(k{\gamma }_k-1)$ and $M_2(T^{‴})\le (n-k{\gamma }_k)[n-(k-1){\gamma }_k]+(4k-2){\gamma }_k-4$ with equality if and only if $T^{‴}\cong T_{n,k,{\gamma }_k}$ by induction on ${\gamma }_k$. The statement is true for ${\gamma }_k=3$ and $n\ge (k+1){\gamma }_k$ by Lemma 3.3. Assume that ${\gamma }_k\ge 4$, the statement holds for ${\gamma }_k-1$ and all the $n\ge (k+1)({\gamma }_k-1)$.

In view of ${\gamma }_k(T^{‴}-\{v_0,v_1,\dots ,v_k\})={\gamma }_k-1$ and $|V(T^{‴}-\{v_0,v_1,\dots ,v_k\})|=n-k-1\ge (k+1)({\gamma }_k-1)$, by the induction hypothesis, we get

$$\begin{array}{rl}{M}_1\left(T^{‴}\right)& =M_1(T^{‴}-\{v_0,v_1,\dots ,v_k\})+2d_{T^{‴}}(v_{k+1})-1+\sum _{i=0}^k{d}_{T^{‴}}^2(v_i)\\ & \le M_1(T_{n-k-1,k,{\gamma }_k-1})+2(n-k{\gamma }_k)+4k\\ & =(n-k{\gamma }_k)(n-k{\gamma }_k+1)+4(k{\gamma }_k-1).\end{array}$$

The equality holds if and only if $T^{‴}-\{v_0,v_1,\dots ,v_k\}\cong T_{n-k-1,k,{\gamma }_k-1}$ and $d_{T^{‴}}(v_{k+1})=\mathrm{△}=n-k{\gamma }_k$. Recalling that $d_{T^{‴}}(v_i)=2$ for $i=1,\dots ,k$, we have $M_1(T^{‴})=(n-k{\gamma }_k)(n-k{\gamma }_k+1)+4(k{\gamma }_k-1)$ if and only if $T^{‴}\cong T_{n,k,{\gamma }_k}$.

Thus, $M_1(T)\le M_1(T^{\prime })\le M_1(T^{″})\le M_1(T^{‴})\le (n-k{\gamma }_k)(n-k{\gamma }_k+1)+4(k{\gamma }_k-1)$ and $M_1(T)=(n-k{\gamma }_k)(n-k{\gamma }_k+1)+4(k{\gamma }_k-1)$ if and only if at least one of the following conditions holds:

1. $T\cong T^{\prime }\cong T^{″}\cong T^{‴}\cong T_{n,k,{\gamma }_k}$;

2. $T\cong T^{\prime }\cong T^{″}$, where $d_{T^{″}}(v_{k+1})=2$. Besides, $T^{‴}\cong T_{n,k,{\gamma }_k}$.

However, the second condition is impossible. If $T^{‴}\cong T_{n,k,{\gamma }_k}$, then $d_{T^{‴}}(v_{k+1})=n-k{\gamma }_k$ and the number of the pendent vertices in $N_{T^{‴}}(v_{k+1})$ is $n-(k+1){\gamma }_k$. By the definition of $T^{‴}$, we have

$$n-(k+1){\gamma }_k\ge |N_{T^{″}}(v_{i_0})\setminus \{v_{i_0-1},v_{i_0+1}\}|.$$

Hence,

$$\begin{array}{rl}{d}_{T^{″}}(v_{k+1})& =d_{T^{‴}}(v_{k+1})-|N_{T^{″}}(v_{i_0})\setminus \{v_{i_0-1},v_{i_0+1}\}|\\ & \ge d_{T^{‴}}(v_{k+1})-[n-(k+1){\gamma }_k]\\ & ={\gamma }_k\ge 3,\end{array}$$

a contradiction to $d_{T^{″}}(v_{k+1})=2$. Therefore,

$$M_1(T)\le (n-k{\gamma }_k)(n-k{\gamma }_k+1)+4(k{\gamma }_k-1)$$

with equality if and only if $T\cong T_{n,k,{\gamma }_k}$.

Note that ${\gamma }_k(T^{‴}-\{v_0,\dots ,v_{k-1}\})={\gamma }_k-1$ and $|V(T^{‴}-\{v_0,\dots ,v_{k-1}\})|>(k+1)({\gamma }_k-1)$. Then

$$\begin{array}{rl}{M}_2\left(T^{‴}\right)& =M_2(T^{‴}-\{v_0,v_1,\dots ,v_{k-1}\})+d_{T^{‴}}(v_{k+1})+4(k-1)+2\\ & \le M_2(T_{n-k,k,{\gamma }_k-1})+n-k{\gamma }_k+4(k-1)+2\\ & =(n-k{\gamma }_k)[n-(k-1){\gamma }_k]+(4k-2){\gamma }_k-4.\end{array}$$

The equality holds if and only if $T^{‴}-\{v_0,\dots ,v_{k-1}\}\cong T_{n-k,k,{\gamma }_k-1}$ and $d_{T^{‴}}(v_{k+1})=\mathrm{△}=n-k{\gamma }_k$. In consideration of $d_{T^{‴}}(v_i)=2$ for $i=1,\dots ,k$, the equality holds if and only if $T^{‴}\cong T_{n,k,{\gamma }_k}$.

Hence, if $i_1\ne 1$, then $M_2(T)\le M_2(T^{\prime })\le M_2({\tilde{T}}^{″})\le M_2(T^{‴})\le (n-k{\gamma }_k)[n-(k-1){\gamma }_k]+(4k-2){\gamma }_k-4$, with equality if and only if at least one of the following conditions holds:

1. $T\cong T^{\prime }\cong {\tilde{T}}^{″}\cong T^{‴}\cong T_{n,k,{\gamma }_k}$;

2. $T\cong T^{\prime }\cong {\tilde{T}}^{″}$, where $d_{{\tilde{T}}^{″}}(v_k)=d_{{\tilde{T}}^{″}}(v_{k+1})=d_{{\tilde{T}}^{″}}(v_{k+2})=2$ and ${\tilde{T}}^{‴}\cong T_{n,k,{\gamma }_k}$.

Analogous to the analysis of the first Zagreb index, the second condition above is impossible. Thus,

$$M_2(T)\le (n-k{\gamma }_k)[n-(k-1){\gamma }_k]+(4k-2){\gamma }_k-4$$

and the equality holds if and only if $T\cong T_{n,k,{\gamma }_k}$.

Besides, if $i=1$, then $M_2(T)\le (n-k{\gamma }_k)[n-(k-1){\gamma }_k]+(4k-2){\gamma }_k-4$ with equality if and only if $T\cong T_{n,k,{\gamma }_k}$ immediately. This completes the proof. □

Remark 3.5

Borovićanin and Furtula [1] proved

$$M_1(T)\le (n-\gamma )(n-\gamma +1)+4(\gamma -1)$$

and

$$M_2(T)\le 2(n-\gamma +1)(\gamma -1)+(n-\gamma )(n-2\gamma +1),$$

with equality if and only if $T\cong T_{n,\gamma }$, where $T_{n,\gamma }$ is the tree obtained from the star $K_{1,n-\gamma }$ by attaching a pendent edge to its $\gamma -1$ pendent vertices. In this paper, we determine the extremal values on the Zagreb indices of trees with distance k-domination number for $k\ge 2$. Note that the domination number is the special case of the distance k-domination number for $k=1$ and $T_{n,k,{\gamma }_k}\cong T_{n,\gamma }$, $T_{n,k,2}^i\cong T_{n,\gamma }$, $i\in \{1,\dots ,k\}$, when $k=1$. Let T be an n-vertex tree with distance k-domination number ${\gamma }_k$. Then, by using Theorems 3.1, 3.2 and 3.4 and the results in [1], we have

$$M_1(T)\le \{\begin{array}{cc}n(n-1)\hfill & \text{if }{\gamma }_k=1,\hfill \\ (n-k{\gamma }_k)(n-k{\gamma }_k+1)+4(k{\gamma }_k-1)\hfill & \text{if }{\gamma }_k\ge 2,\hfill \end{array}$$

with equality if and only if $T\cong S_n$ when ${\gamma }_k=1$, $T\cong T_{n,k,2}^i$, $i\in \{1,\dots ,k\}$, when ${\gamma }_k=2$, or $T\cong T_{n,k,{\gamma }_k}$ when ${\gamma }_k\ge 3$. Moreover,

$$M_2(T)\le \{\begin{array}{cc}2(n-{\gamma }_k+1)({\gamma }_k-1)+(n-{\gamma }_k)(n-2{\gamma }_k+1)\hfill & \text{if }k=1,\hfill \\ {(n-1)}^2\hfill & \text{if }k\ge 2,{\gamma }_k=1,\hfill \\ (n-k{\gamma }_k)[n-(k-1){\gamma }_k]+(4k-2){\gamma }_k-4\hfill & \text{if }k\ge 2,{\gamma }_k\ge 2,\hfill \end{array}$$

with equality if and only if $T\cong S_n$ when $k\ge 2$ and ${\gamma }_k=1$, $T\cong T_{n,k,2}^i$, $i\in \{2,\dots ,k\}$, when $k\ge 2$ and ${\gamma }_k=2$, or $T\cong T_{n,k,{\gamma }_k}$ otherwise.

Appendix

Proof

Either $M_i(G_1)>M_i(G)$ or $M_i(G_2)>M_i(G)$, $i=1,2$, in Lemma 2.8, where $G_1=G-\{v{y}_1\}+\{u{y}_1\}$ and $G_2=G-\{u{x}_1\}+\{v{x}_1\}$, as shown in the following figure. Let $G^{\ast }=G-\{x_1,\dots ,x_r,y_1,\dots ,y_t\}$, $d_{G^{\ast }}(u)=a$ and $d_{G^{\ast }}(v)=b$. Then

$$\begin{array}{rl}{M}_1(G_1)-M_1(G)& ={(a+r+1)}^2+{(b+t-1)}^2-{(a+r)}^2-{(b+t)}^2\\ & =2(a+r-b-t+1)\end{array}$$

and

$$\begin{array}{rl}{M}_1(G_2)-M_1(G)& ={(a+r-1)}^2+{(b+t+1)}^2-{(a+r)}^2-{(b+t)}^2\\ & =2(b+t-a-r+1)\end{array}$$

by the definition of the first Zagreb index. Suppose that $M_1(G_1)-M_1(G)\le 0$. Then $a+r\le b+t-1$. It follows that $M_1(G_2)-M_1(G)>0$.

If $u\notin N_G(v)$, then

$$\begin{array}{rl}{M}_2(G_1)-M_2(G)=& (a+r+1)(\sum _{x\in N_{G^{\ast }}(u)}{d}_G(x)+r+1)\\ & +(b+t-1)(\sum _{x\in N_{G^{\ast }}(v)}{d}_G(x)+t-1)\\ & -(a+r)(\sum _{x\in N_{G^{\ast }}(u)}{d}_G(x)+r)-(b+t)(\sum _{x\in N_{G^{\ast }}(v)}{d}_G(x)+t)\\ =& \sum _{x\in N_{G^{\ast }}(u)}{d}_G(x)-\sum _{x\in N_{G^{\ast }}(v)}{d}_G(x)+2r-2t+a-b+2\end{array}$$

and

$$\begin{array}{rl}{M}_2(G_2)-M_2(G)=& (a+r-1)(\sum _{x\in N_{G^{\ast }}(u)}{d}_G(x)+r-1)\\ & +(b+t+1)(\sum _{x\in N_{G^{\ast }}(v)}{d}_G(x)+t+1)\\ & -(a+r)(\sum _{x\in N_{G^{\ast }}(u)}{d}_G(x)+r)-(b+t)(\sum _{x\in N_{G^{\ast }}(v)}{d}_G(x)+t)\\ =& \sum _{x\in N_{G^{\ast }}(v)}{d}_G(x)-\sum _{x\in N_{G^{\ast }}(u)}{d}_G(x)+2t-2r+b-a+2.\end{array}$$

If $M_2(G_1)-M_2(G)\le 0$, then $M_2(G_2)-M_2(G)>0$.

If $u\in N_G(v)$, then

$$\begin{array}{c}{M}_2(G_1)-M_2(G)\\ =(a+r+1)({\sum }_{x\in N_{G^{\ast }}(u)\setminus \{v\}}{d}_G(x)+r+1)+(b+t-1)({\sum }_{x\in N_{G^{\ast }}(u)\setminus \{v\}}{d}_G(x)+t-1)\\ +(a+r+1)(b+t-1)-(a+r)({\sum }_{x\in N_{G^{\ast }}(u)\setminus \{v\}}{d}_G(x)+r)\\ -(b+t)({\sum }_{x\in N_{G^{\ast }}(u)\setminus \{v\}}{d}_G(x)+t)-(a+r)(b+t)\\ ={\sum }_{x\in N_{G^{\ast }}(u)\setminus \{v\}}{d}_G(x)-{\sum }_{x\in N_{G^{\ast }}(v)\setminus \{u\}}{d}_G(x)+r-t+1\end{array}$$

and

$$\begin{array}{c}{M}_2(G_2)-M_2(G)\\ =(a+r-1)({\sum }_{x\in N_{G^{\ast }}(u)\setminus \{v\}}{d}_G(x)+r-1)+(b+t+1)({\sum }_{x\in N_{G^{\ast }}(u)\setminus \{v\}}{d}_G(x)+t+1)\\ +(a+r-1)(b+t+1)-(a+r)({\sum }_{x\in N_{G^{\ast }}(u)\setminus \{v\}}{d}_G(x)+r)\\ -(b+t)({\sum }_{x\in N_{G^{\ast }}(u)\setminus \{v\}}{d}_G(x)+t)-(a+r)(b+t)\\ ={\sum }_{x\in N_{G^{\ast }}(v)\setminus \{u\}}{d}_G(x)-{\sum }_{x\in N_{G^{\ast }}(u)\setminus \{v\}}{d}_G(x)+t-r+1.\end{array}$$

Assume that $M_2(G_1)-M_2(G)\le 0$. Then $M_2(G_2)-M_2(G)>0$. Therefore, either $M_i(G_1)>M_i(G)$ or $M_i(G_2)>M_i(G)$, $i=1,2$. □

Authors’ contributions

All authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.