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

Journal of Inequalities and Applications20182018:16

https://doi.org/10.1186/s13660-017-1597-3

  • Received: 25 September 2017
  • Accepted: 15 December 2017
  • Published:

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 \(u\in V(G)\setminus D\), \(d_{G}(u,v)\leq k\) 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.

Keywords

  • first Zagreb index
  • second Zagreb index
  • trees
  • distance k-domination number

MSC

  • 05C35
  • 05C69

1 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 \(\varepsilon_{G}(v)=\max\{ d_{G}(u,v)\mid u\in V(G)\}\). The diameter of G is \(\operatorname{diam}(G)=\max\{ \varepsilon_{G}(v)\mid v\in V(G)\}\). A path P is called a diameter path of G if the length of P is \(\operatorname{diam}(G)\). 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 \(\bigcup_{i=0}^{k}N^{i}_{G}(v)\cap D=\{u\}\). That is, \(d_{G}(v,u)\leq k\) and \(d_{G}(v,x)\geq k+1\) 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 [1113], 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 [2025]. Many researchers are interested in establishing the bounds for the Zagreb indices of graphs and characterizing the extremal graphs [1, 2640].

A set \(D\subseteq V(G)\) is a dominating set of G if, for any vertex \(u\in V(G)\setminus D\), \(N_{G}(u)\cap D\ne\emptyset\). 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 \(u\in V(G)\setminus D\), \(d_{G}(u,v)\leq k\) 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, 4547].

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.

2 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 \(T'\) 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 \(T'=\tau(T,uv)\).
Figure 1
Figure 1

T and \(\pmb{T'}\) in Transformation I .

Lemma 2.2

Let T be a tree of order n (≥3) and \(T'=\tau(T,uv)\). Then \(M_{i}(T')>M_{i}(T)\), \(i=1,2\).

Proof

It is obvious that \(d_{T'}(u)=d_{T}(u)+d_{T}(v)-1\) and
$$\begin{aligned} M_{1}\bigl(T'\bigr)-M_{1}(T)={}& \bigl(d_{T}(u)+d_{T}(v)-1\bigr)^{2}+1-d_{T}^{2}(u)-d_{T}^{2}(v) \\ ={}&2\bigl(d_{T}(u)-1\bigr) \bigl(d_{T}(v)-1\bigr) \\ >{}&0.\end{aligned} $$
Let \(x\in V(T)\) be a vertex different from u and v. Then
$$\begin{aligned} M_{2}\bigl(T'\bigr)-M_{2}(T)={}& \bigl(d_{T}(u)+d_{T}(v)-1\bigr) \biggl(\sum _{xu\in E(T_{1})}d_{T}(x)+\sum_{xv\in E(T_{2})}d_{T}(x)+1 \biggr) \\ &-d_{T}(u)\sum_{xu\in E(T_{1})}d_{T}(x)-d_{T}(v) \sum_{xv\in E(T_{2})}d_{T}(x) -d_{T}(u)d_{T}(v) \\ ={}&\bigl(d_{T}(v)-1\bigr)\sum_{xu\in E(T_{1})}d_{T}(x)+ \bigl(d_{T}(u)-1\bigr)\sum_{xv\in E(T_{2})}d_{T}(x) \\ &+d_{T}(u)+d_{T}(v)-1-d_{T}(u)d_{T}(v) \\ \geq{} &2\bigl(d_{T}(v)-1\bigr) \bigl(d_{T}(u)-1 \bigr)+d_{T}(u)+d_{T}(v)-1-d_{T}(u)d_{T}(v) \\ ={}&\bigl(d_{T}(v)-1\bigr) \bigl(d_{T}(u)-1\bigr) \\ >{}&0.\end{aligned} $$
This completes the proof. □

Lemma 2.3

([50])

Let u and v be two distinct vertices in G. \(u_{1},u_{2},\ldots,u_{r}\) are the pendent vertices adjacent to u and \(v_{1}, v_{2},\ldots,v_{t}\) are the pendent vertices adjacent to v. Define \(G'=G-\{vv_{1}, vv_{2},\ldots, vv_{t}\}+ \{uv_{1}, uv_{2},\ldots, uv_{t}\}\) and \(G''=G-\{uu_{1}, uu_{2},\ldots, uu_{r}\}+\{ vu_{1}, vu_{2},\ldots, vu_{r}\}\), as shown in Figure 2. Then either \(M_{i}(G')>M_{i}(G)\) or \(M_{i}(G'')>M_{i}(G)\), \(i=1,2\).
Figure 2
Figure 2

G , \(\pmb{G'}\) and \(\pmb{G''}\) in Lemma  2.3 .

Lemma 2.4

([51])

For a connected graph G of order n with \(n\geq k+1\), \(\gamma_{k}(G)\leq\lfloor\frac {n}{k+1}\rfloor\).

Let G be a connected graph of order n. If \(\gamma_{k}(G)\geq2\), then \(n\geq k+1\). Otherwise, \(\gamma_{k}(G)=1\), a contradiction. Hence, by Lemma 2.4, we have \(\gamma_{k}(G)\leq\lfloor\frac{n}{k+1}\rfloor\) and \(n\geq(k+1)\gamma_{k}\) for any connected graph G of order n if \(\gamma _{k}(G)\geq2\).

Lemma 2.5

Let T be an n-vertex tree with distance k-domination number \(\gamma_{k}\geq2\). Then \(\triangle \leq n-k\gamma_{k}\).

Proof

Suppose that \(\triangle\geq n-k\gamma_{k}+1\). Let \(v\in V(T)\) be the vertex such that \(d(v)=\triangle\) and \(N(v)=\{v_{1},\ldots ,v_{\triangle}\}\). Denote by \(T^{i}\) the component of \(T-v\) containing the vertex \(v_{i}\), \(i=1,\ldots,\triangle\). Let D be a minimum distance k-dominating set of T,
$$S_{1}=\bigl\{ i\mid i\in\{1,2,\ldots,\triangle\}, 0\leq\varepsilon _{T^{i}}(v_{i})\leq k-1\bigr\} $$
and
$$S_{2}=\bigl\{ i\mid i\in\{1,2,\ldots,\triangle\}, \varepsilon _{T^{i}}(v_{i})\geq k\bigr\} . $$
Clearly, \(|S_{2}|\geq1\). If not, \(\{v\}\) is a distance k-dominating set of T, which contradicts \(\gamma_{k}\geq2\). If \(|S_{1}|=0\), then \(\varepsilon_{T^{i}}(v_{i})\geq k\) for \(i=1,\ldots ,\triangle\), so \(|V(T^{i})\cap D|\geq1\). Therefore, \(\gamma_{k}\geq\triangle\geq {n-k\gamma_{k}+1}\), which implies that \(\gamma_{k}\geq\frac{n+1}{k+1}\). Since \(\gamma_{k}\geq2\), \(\gamma_{k}\leq\lfloor\frac{n}{k+1}\rfloor\) by Lemma 2.4, a contradiction. Thus, \(|S_{1}|\geq1\). Let \(i_{1}\in S_{1}\) and
$$\varepsilon_{T^{i_{1}}}(v_{i_{1}})=\max\bigl\{ \varepsilon_{T^{i}}(v_{i}) \mid i\in S_{1}\bigr\} =\lambda. $$
Then \(0\leq\lambda\leq k-1\), so \(|S_{2}|\leq\lfloor\frac{n-\triangle -1-\lambda}{k}\rfloor \leq\lfloor\frac{k\gamma_{k}-2}{k}\rfloor\leq\gamma_{k}-1\).

If \(V(T^{i})\cap D=D_{1}\neq\emptyset\) 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 \(V(T^{i})\cap D=\emptyset\) for each \(i\in S_{1}\). Similarly, suppose that \(D'\cap V(T^{i_{1}})=\emptyset\) where \(D'\) is a minimum distance k-dominating set of the tree \(T'=T-\bigcup_{i\in S_{1}\setminus\{i_{1}\}}V(T^{i})\).

We claim that \(D'\) 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 \(y'\in D'_{1}=\bigcup_{i=0}^{k}N_{T'}^{i}(y)\cap D'\). Then \(y'\in V(T')\setminus V(T^{i_{1}})\) and \(d(y,y')=d(y,v)+d(v,y')\leq k\), so, for \(x\in\bigcup_{i\in S_{1}\setminus\{i_{1}\}}V(T^{i})\), we have \(d(x,y')=d(x,v)+d(v,y')\leq d(y,v)+d(v,y')\leq k\). Hence, all the vertices in \(\bigcup_{i\in S_{1}\setminus\{i_{1}\}}V(T^{i})\) can be dominated by \(y'\in D'\). Therefore, \(D'\) is a distance k-dominating set of T, so the claim is true.

In view of
$$k+1< (k+1)|S_{2}|+\lambda+2\leq\big|V\bigl(T'\bigr)\big|\leq n-|S_{1}|+1=n-\triangle+|S_{2}|+1, $$
one has
$$\begin{aligned} \gamma_{k}&\leq\big|D'\big| \\ &\leq\biggl\lfloor \frac{n-\triangle+|S_{2}|+1}{k+1}\biggr\rfloor \quad (\text{by Lemma 2.4}) \\ &\leq\biggl\lfloor \frac{(k+1)\gamma_{k}-1}{k+1}\biggr\rfloor \quad\bigl(\text{since }\triangle \geq n-k\gamma_{k}+1, |S_{2}|\leq\gamma_{k}-1\bigr) \\ &< \gamma_{k},\end{aligned} $$
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 \(\gamma_{k}\geq2\), then \(\gamma_{k}(T)\leq\frac{n-\Delta(T)}{k}\).

Let \(\mathscr{T}_{n,k,\gamma_{k}}\) 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 \(v_{1},v_{2},\ldots,v_{n-k\gamma_{k}}\). 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 \(i\in\{2,\ldots,\gamma_{k}\}\), as shown in Figure 3. Then \(T_{n,k,\gamma_{k}}\in\mathscr{T}_{n,k,\gamma_{k}}\). Even more noteworthy is the notion that \(\gamma_{k}(T_{n,k,\gamma _{k}})=\gamma_{k}=\frac{n-\Delta(T_{n,k,\gamma_{k}})}{k}\). It implies that the upper bound on the distance k-domination number mentioned in the above paragraph is sharp.
Figure 3
Figure 3

\(\pmb{T_{n,k,\gamma_{k}}}\) .

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
$$M_{2}(T_{n,k,\gamma_{k}})= \textstyle\begin{cases} (n-k\gamma_{k})[n-(k-1)\gamma_{k}]+(4k-2)\gamma_{k}-4& \text{if }k\geq2,\\ 2(n-\gamma+1)(\gamma-1)+(n-\gamma)(n-2\gamma+1) &\text{if }k=1. \end{cases} $$
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 \(k\geq2\) 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. (1)

    T is any tree on \(k+1\) vertices;

     
  2. (2)

    \(T=R\circ k\) for some tree R on \(n\geq1\) 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 \(\gamma_{k}(T)\geq3\). If \(n=(k+1)\gamma _{k}\), then
$$M_{1}(T)\leq\gamma_{k}(\gamma_{k}+1)+4(k \gamma_{k}-1) $$
and
$$M_{2}(T)\leq2\gamma_{k}^{2}+(4k-2) \gamma_{k}-4, $$
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 \(V(R)=\{v_{1},\ldots,v_{\gamma_{k}}\}\). 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 \(\{u_{1},\ldots ,u_{n}\}\). Hence, \(\sum_{i=1}^{\gamma_{k}}d_{R}(v_{i})=2(\gamma_{k}-1)\). By the definition of the first Zagreb index, we have
$$\begin{aligned} M_{1}(T) =&\sum_{i=1}^{\gamma_{k}}d_{T}^{2}(v_{i})+ \sum_{x\in V(T)\setminus V(R)}d_{T}^{2}(x) \\ =&\sum_{i=1}^{\gamma_{k}}\bigl(d_{T}(v_{i})-1 \bigr)^{2}+\sum_{x\in V(T)\setminus V(R)}d_{T}^{2}(x)+2 \sum_{i=1}^{\gamma_{k}}\bigl(d_{T}(v_{i})-1 \bigr)+\gamma_{k} \\ =&M_{1}(R)+4(k-1)\gamma_{k}+\gamma_{k}+2\sum _{i=1}^{\gamma_{k}}d_{R}(v_{i})+ \gamma _{k} \\ \leq&M_{1}(S_{\gamma_{k}})+4(k-1)\gamma_{k}+2 \gamma_{k}+4(\gamma_{k}-1) \\ =&\gamma_{k}(\gamma_{k}+1)+4(k\gamma_{k}-1). \end{aligned}$$
The equality holds if and only if \(R\cong S_{\gamma_{k}}\), that is, \(T\cong T_{n,k,\gamma_{k}}\). We have
$$\begin{aligned} M_{2}(T)={}&\sum_{xy\in E(R)}d_{T}(x)d_{T}(y)+ \sum_{xy\in E(T)\setminus E(R)}d_{T}(x)d_{T}(y) \\ ={}&\sum_{xy\in E(R)}\bigl(d_{T}(x)-1\bigr) \bigl(d_{T}(y)-1\bigr)+\sum_{xy\in E(R)} \bigl(d_{T}(x)+d_{T}(y)-1\bigr) \\ &+\sum_{xy\in E(T)\setminus E(R)}d_{T}(x)d_{T}(y) \\ ={}&M_{2}(R)+\sum_{x\in V(R)}d_{T}(x) \bigl(d_{T}(x)-1\bigr)-(\gamma_{k}-1) \\ &+\sum_{x\in V(R)}2d_{T}(x)+4(k-2) \gamma_{k}+2\gamma_{k} \\ ={}&M_{2}(R)+\sum_{x\in V(R)} \bigl(d_{T}(x)-1\bigr)^{2} +3\sum _{x\in V(R)}\bigl(d_{T}(x)-1\bigr)+4k\gamma_{k}-5 \gamma_{k}-1 \\ ={}&M_{2}(R)+M_{1}(R)+6(\gamma_{k}-1)+4k \gamma_{k}-5\gamma_{k}+1 \\ \leq{}&M_{2}(S_{\gamma_{k}})+M_{1}(S_{\gamma_{k}})+4k \gamma_{k}+\gamma_{k}-5 \\ ={}&2\gamma_{k}^{2}+(4k-2)\gamma_{k}-4. \end{aligned} $$
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 \(S_{G}\neq\emptyset\), then \(|N_{G}(S_{G})|=1\).

Proof

Suppose that \(|N_{G}(S_{G})|\geq2\) and u and v are two distinct vertices in \(N_{G}(S_{G})\). \(x_{1},x_{2},\ldots, x_{r}\) are the pendent vertices adjacent to u and \(y_{1},y_{2},\ldots,y_{t}\) are the pendent vertices adjacent to v, where \(r\geq1\) and \(t\geq1\). Let D be a minimum distance k-dominating set of G. If \(x_{i}\in D\) for some \(i\in\{1,\ldots,r\}\), then \(D-x_{i}+u\) is a distance k-dominating set of T. Hence, we assume that \(x_{i}\notin D\), \(i=1,\ldots,r\). Similarly, \(y_{i}\notin D\) for \(1\leq i\leq t\). Define \(G_{1}=G-\{vy_{1}\}+\{uy_{1}\}\) and \(G_{2}=G-\{ux_{1}\}+\{vx_{1}\}\). 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. □

3 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 \(v_{i}\), \(i=1,2,\ldots,d-1\). 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 \(M_{1}(T)\leq n(n-1)\) and \(M_{2}(T)\leq(n-1)^{2}\). The equality holds if and only if \(T\cong S_{n}\).

Let \(T^{i}_{n,k,2}\) 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 \(i\in\{1,\ldots,2k\}\).

Theorem 3.2

If T is an n-vertex tree with distance k-domination number \(\gamma_{k}(T)=2\), then
$$M_{1}(T)\leq(n-2k) (n-2k+1)+4(2k-1), $$
with equality if and only if \(T\cong T^{i}_{n,k,2}\), where \(i\in\{1,\ldots,k\}\). Also,
$$M_{2}(T)\leq(n-2k) (n-2k+2)+8k-8, $$
with equality if and only if \(T\cong T^{i}_{n,k,2}\), where \(i\in\{2,\ldots,k\}\).

Proof

Assume that \(T\in\mathscr{T}_{n,k,2}\) is the tree that maximizes the Zagreb indices and \(P=v_{0}v_{1}\cdots v_{d}\) is a diameter path of T. If \(d\leq2k\), then \(\{v_{\lfloor\frac{d}{2}\rfloor}\}\) is a distance k-dominating set of T, a contradiction to \(\gamma_{k}(T)=2\). If \(d\geq2k+2\), define \(T'=\tau(T,v_{i}v_{i+1})\), where \(i\in\{1,\ldots ,d-2\}\). Then \(T'\in\mathscr{T}_{n,k,2}\). By Lemma 2.2, we have \(M_{i}(T')>M_{i}(T)\), \(i=1,2\), a contradiction. Hence, \(d=2k+1\).

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

Since \(M_{1}(T^{i}_{n,k,2})=M_{1}(T^{j}_{n,k,2})\) for \(1\leq i\neq j\leq d-1\) and \(T^{i}_{n,k,2}\cong T^{d-i}_{n,k,2}\) for \(k+1\leq i\leq d-1\), we get \(T\cong T^{i}_{n,k,2}\), \(i\in\{1,\ldots,k\}\). By direct computation, one has \(M_{1}(T)=M_{1}(T^{i}_{n,k,2})= (n-2k)(n-2k+1)+4(2k-1)\), \(i\in\{1,\ldots ,k\}\). In addition, \(M_{2}(T^{1}_{n,k,2})=M_{2}(T^{d-1}_{n,k,2})< M_{2}(T^{2}_{n,k,2})=\cdots =M_{2}(T^{d-2}_{n,k,2})\) and \(T^{i}_{n,k,2}\cong T^{d-i}_{n,k,2}\) for \(i\in \{k+1,\ldots,d-2\}\). Hence, \(T\cong T^{i}_{n,k,2}\), where \(i\in\{2,\ldots,k\}\). Moreover, \(M_{2}(T)=M_{2}(T^{i}_{n,k,2})=(n-2k)(n-2k+2)+8k-8\). This completes the proof. □

Lemma 3.3

Let tree \(T\in\mathscr{T}_{n,k,3}\). Then
$$M_{1}(T)\leq(n-3k) (n-3k+1)+4(3k-1) $$
and
$$M_{2}(T)\leq(n-3k) (n-3k+3)+12k-10, $$
with equality if and only if \(T\cong T_{n,k,3}\).

Proof

Assume that \(T\in\mathscr{T}_{n,k,3}\). We complete the proof by induction on n. By Lemma 2.4, we have \(n\geq (k+1)\gamma_{k}\). 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 \(d\geq2k+2\). Otherwise, \(\{v_{k},v_{k+1}\}\) is a distance k-dominating set, a contradiction. Note that \(\bigcup_{i=0}^{k}N_{T}^{i}(v_{0})\cap D\neq\emptyset\) and \(\bigcup_{i=0}^{k}N_{T}^{i}(v_{0})\subseteq(\bigcup_{i=0}^{k-1}V(T_{i})\cup\{v_{k}\})\). Hence, \((\bigcup_{i=0}^{k-1}V(T_{i})\cup\{v_{k}\})\cap D\neq\emptyset\). 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 \((\bigcup_{i=0}^{k}V(T_{i})\setminus\{ v_{k}\})\cap D=\emptyset\). Similarly, \(v_{d-k}\in D\) and \((\bigcup_{i=d-k}^{d}V(T_{i})\setminus\{v_{d-k}\})\cap D=\emptyset\). Suppose that \(v_{0}=u_{1}\), \(v_{d}=u_{2},\ldots, u_{m}\) are the pendent vertices of T and \(S_{T}=\{u_{i}\mid1\leq i\leq m, \gamma_{k}(T-u_{i})=\gamma_{k}(T)\}\). We have the following claim.

Claim 1

\(S_{T}\neq\emptyset\).

Proof

Assume that \(S_{T}=\emptyset\). Namely, \(\gamma_{k}(T-u_{i})=\gamma_{k}(T)-1\) for each \(i\in\{1,\ldots,m\}\). If \(D\setminus\{w_{i}\}\) is a minimum distance k-dominating set of the tree \(T-u_{i}\), where \(w_{i}\in D\), then \(w_{i}\neq w_{j}\) for \(1\leq i\neq j\leq m\). Otherwise, \(\gamma_{k}(T-u_{i})=\gamma_{k}(T)\) or \(\gamma _{k}(T-u_{j})=\gamma_{k}(T)\), a contradiction. It follows that \(m\leq\gamma_{k}\).

If \(d_{T}(v_{i})\geq3\) for some \(i\in\{2,\ldots,k,d-k,\ldots,d-1\}\), then \(V(T_{i})\cap\{u_{3},\ldots,u_{m}\}\neq\emptyset\). In view of \(\{ v_{k},v_{d-k}\}\subseteq D\), we have \(\gamma_{k}(T-x)=\gamma_{k}(T)\) for \(x\in V(T_{i})\cap\{u_{3},\ldots,u_{m}\}\), a contradiction. Hence, \(d_{T}(v_{i})=2\) for \(i\in\{2,\ldots,k,d-k,\ldots,d-1\}\).

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 \((\bigcup_{i=0}^{k}V(T_{i})\setminus\{v_{k}\})\cap D=\emptyset\), 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 \(d_{T}(v_{k+1})\geq3\). Hence, \(m\geq3=\gamma_{k}\). Recalling that \(m\leq \gamma_{k}=3\), 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 \(S_{T}\neq\emptyset\) for \(T\in\mathscr{T}_{n,k,3}\), the tree among \(\mathscr{T}_{n,k,3}\) that maximizes the Zagreb indices must be in the set \(\{T^{*}\in\mathscr{T}_{n,k,3}\mid |N_{T^{*}}(S_{T^{*}})|=1\}\) by Lemma 2.8. To determine the extremal trees among \(\mathscr{T}_{n,k,3}\), we assume that \(T\in\{T^{*}\in\mathscr {T}_{n,k,3}\mid|N_{T^{*}}(S_{T^{*}})|=1\}\) 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, \(d_{T}(s)\leq\triangle\leq n-k\gamma_{k}\). Define \(A=\{x\in V(T)\mid d_{T}(x)=1, xs\notin E(T)\}\) and \(B=\{x\in V(T)\mid d_{T}(x)\geq2, xs\notin E(T)\}\). Then \(\gamma_{k}(T-x)=\gamma _{k}(T)-1\) for all \(x\in A\). As a consequence, \(|A|\leq\gamma_{k}\) from the proof of Claim 1. By the induction hypothesis,
$$\begin{aligned} M_{1}(T) =& M_{1}(T-u_{i})+2d(s) \\ \leq&(n-1-k\gamma_{k}) (n-1-k\gamma_{k}+1)+4(k \gamma_{k}-1)+2(n-k\gamma _{k}) \\ =&(n-k\gamma_{k}) (n-k\gamma_{k}+1)+4(k \gamma_{k}-1). \end{aligned}$$
The equality holds if and only if \(T-u_{i}\cong T_{n-1,k,\gamma_{k}}\) and \(d_{T}(s)=\triangle=n-k\gamma_{k}\), i.e., \(T\cong T_{n,k,\gamma_{k}}\).
Note that \(|A|+|B|=n-1-d_{T}(s)\) and \(|A|\leq\gamma_{k}\). Therefore, \(|B|=n-1-d_{T}(s)-|A|\geq n-1-d_{T}(s)-\gamma_{k}\) and
$$\sum_{xs\notin E(T)}d(x)\geq|A|+2|B|=\bigl(|A|+|B|\bigr)+|B|\geq 2 \bigl(n-1-d_{T}(s)\bigr)-\gamma_{k}. $$
By the above inequality and the definition of \(M_{2}\), we have
$$\begin{aligned} M_{2}(T) =&M_{2}(T-u_{i})+\sum _{v\in V(T)}d_{T}(v)-\sum_{xs\notin E(T)}d_{T}(x)-1 \\ \leq&M_{2}(T-u_{i})+2(n-1)-2\bigl(n-1-d_{T}(s) \bigr)+\gamma_{k}-1 \end{aligned}$$
(1)
$$\begin{aligned} \leq&(n-1-k\gamma_{k})\bigl[n-1-(k-1)\gamma_{k} \bigr]+(4k-2)\gamma_{k}-4 \\ &{}+2(n-k\gamma_{k})+\gamma_{k}-1 \quad\bigl(\text{since } d_{T}(s)\leq\triangle\leq n-k\gamma_{k}\bigr) \\ =&(n-k\gamma_{k})\bigl[n-(k-1)\gamma_{k}\bigr]+(4k-2) \gamma_{k}-4. \end{aligned}$$
(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 \(d_{T}(s)=\triangle=n-k\gamma_{k}\). Hence, \(M_{2}(T)\leq(n-k\gamma _{k})[n-(k-1)\gamma_{k}]+(4k-2)\gamma_{k}-4\) 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
$$M_{1}(T)\leq(n-k\gamma_{k}) (n-k\gamma_{k}+1)+4(k \gamma_{k}-1) $$
and
$$M_{2}(T)\leq(n-k\gamma_{k})\bigl[n-(k-1)\gamma_{k} \bigr]+(4k-2)\gamma_{k}-4, $$
with equality if and only if \(T\cong T_{n,k,\gamma_{k}}\).

Proof

Let \(T\in\mathscr{T}_{n,k,\gamma_{k}}\) 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 \(S_{T}=\emptyset\), then \(\gamma_{k}(T-v_{i})=\gamma_{k}(T)-1\) for \(i=0,d\). If \(S_{T}\neq\emptyset\), then we suppose that \(T\in\{T^{*}\in\mathscr {T}_{n,k,\gamma_{k}}\mid|N_{T^{*}}(S_{T^{*}})|=1\}\) by Lemma 2.8 for establishing the maximum Zagreb indices of trees among \(\mathscr {T}_{n,k,\gamma_{k}}\). If \(v_{d}\in S_{T}\neq\emptyset\), 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 \(T'\) be the tree obtained from T by applying Transformation I on \(T_{i}\) repeatedly for \(i=1,\ldots,k\) such that \(T'_{i}\cong S_{|V(T'_{i})|}\), where \(T'_{i}\) is the component of \(T'-\{ v_{i-1}v_{i},v_{i}v_{i+1}\}\) containing \(v_{i}\), \(i=1,\ldots,k\) (see Figure 4). Then \(T'\in\mathscr{T}_{n,k,\gamma_{k}}\). By Lemma 2.2, we have \(M_{i}(T)\leq M_{i}(T')\), \(i=1,2\), with equality if and only if \(T\cong T'\).
Figure 4
Figure 4

T , \(\pmb{T'}\) , \(\pmb{T''}\) and \(\pmb{T'''}\) .

By Lemma 2.3, for some \(i_{0},i_{1}\in\{1,\ldots,k\}\), define
$$\begin{aligned} T''={}&T'-\bigcup _{i\in\{1,\ldots,k\}\setminus\{i_{0}\}}\bigl\{ v_{i} x\mid x\in N_{T'}(v_{i}) \setminus\{v_{i-1},v_{i+1}\}\bigr\} \\ &+\bigcup_{i\in\{1,\ldots,k\}\setminus\{i_{0}\}}\bigl\{ v_{i_{0}}x\mid x\in N_{T'}(v_{i})\setminus\{v_{i-1},v_{i+1}\} \bigr\} \end{aligned} $$
and
$$\begin{aligned} \widetilde{T}''={}&T'-\bigcup _{i\in\{1,\ldots,k\}\setminus\{i_{1}\}}\bigl\{ v_{i} x\mid x\in N_{T'}(v_{i}) \setminus\{v_{i-1},v_{i+1}\}\bigr\} \\ & +\bigcup_{i\in\{1,\ldots,k\}\setminus\{i_{1}\}}\bigl\{ v_{i_{1}}x\mid x\in N_{T'}(v_{i})\setminus\{v_{i-1},v_{i+1}\} \bigr\} .\end{aligned} $$
Then one has \(M_{1}(T')\leq M_{1}(T'')\) with equality if and only if \(T'\cong T''\) and \(M_{2}(T')\leq M_{2}(\widetilde{T}'')\) with equality if and only if \(T'\cong\widetilde{T}''\).
Suppose that \(|N_{T''}(v_{i_{0}})\setminus\{v_{i_{0}-1},v_{i_{0}+1}\} |=|N_{\widetilde{T}''}\setminus\{v_{i_{1}-1},v_{i_{1}+1}\}|=m\), \(m\geq0\). Let
$$\begin{aligned} T'''={}&T''- \bigl\{ v_{i_{0}}x\mid x\in N_{T''}(v_{i_{0}})\setminus\{ v_{i_{0}-1},v_{i_{0}+1}\}\bigr\} \\ &+\bigl\{ v_{k+1}x\mid x\in N_{T''}(v_{i_{0}})\setminus \{v_{i_{0}-1},v_{i_{0}+1}\} \bigr\} \\ ={}&\widetilde{T}''-\bigl\{ v_{i_{1}}x\mid x\in N_{\widetilde {T}''}(v_{i_{1}})\setminus\{v_{i_{1}-1},v_{i_{1}+1}\} \bigr\} \\ &+\bigl\{ v_{k+1}x\mid x\in N_{\widetilde{T}''}(v_{i_{1}})\setminus \{ v_{i_{1}-1},v_{i_{1}+1}\}\bigr\} .\end{aligned} $$
Then D is a minimum distance k-dominating set of \(T'''\) and \(d_{T'''}(v_{i})=2\) for \(i=1,\ldots,k\). 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 \(\bigcup_{i=0}^{k}N_{T'''}^{i}(v_{k})\setminus\{v_{0},\ldots,v_{k}\}\) can be dominated by \(v_{k+1}\in D\). Thus, \(D\setminus\{v_{k}\}\) is a distance k-dominating set of tree \(T'''-\{v_{0},\ldots,v_{k}\}\). In addition, \(\mathit{PN}_{k,D}(v_{k+1})\subseteq V(T''')\setminus\{v_{0},\ldots ,v_{k}\}\), which means that \(D\setminus\{v_{k}\}\) is a minimum distance k-dominating set of \(T'''-\{v_{0},\ldots,v_{k}\}\). So \(\gamma_{k}(T'''-\{v_{0},\ldots,v_{k}\} )=\gamma_{k}-1\). Analogously, \(\gamma_{k}(T'''-\{v_{0},\ldots,v_{k-1}\})=\gamma_{k}-1\).
By the definition of the first Zagreb index, we get
$$\begin{aligned} M_{1}\bigl(T''' \bigr)-M_{1}\bigl(T''\bigr)&=4+ \bigl(d_{T''}(v_{k+1})+m\bigr)^{2}-(2+m)^{2}-d_{T''}^{2}(v_{k+1}) \\ &=2m\bigl(d_{T''}(v_{k+1})-2\bigr) \\ &\geq0,\end{aligned} $$
so \(M_{1}(T''')-M_{1}(T'')=0\) if and only if at least one of the following conditions holds:
  1. (1)

    \(m=0\), which implies that \(T''\cong T'''\);

     
  2. (2)

    \(d_{T''}(v_{k+1})=2\).

     
If \(i_{1}=1\), then
$$\begin{aligned} M_{2}\bigl(T''' \bigr)-M_{2}\bigl(\widetilde{T}''\bigr) ={}&6+ \bigl(d_{\widetilde{T}''}(v_{k+1})+m\bigr) \biggl(m+\sum _{x\in N_{\widetilde {T}''}(v_{k+1})}d_{\widetilde{T}''}(x)\biggr) \\ &-(m+2) (m+3)-d_{\widetilde{T}''}(v_{k+1})\sum _{x\in N_{\widetilde {T}''}(v_{k+1})}d_{\widetilde{T}''}(x) \\ ={}&m\biggl[d_{\widetilde{T}''}(v_{k+1})+\sum_{x\in N_{\widetilde {T}''}(v_{k+1})}d_{\widetilde{T}''}(x)-5 \biggr] \\ \geq{}&0 ,\end{aligned} $$
with equality if and only if \(m=0\), that is, \(\widetilde{T}''\cong T'''\). If \(i_{1}\neq1\) and \(i_{1}\neq k\), then
$$\begin{aligned} M_{2}\bigl(T''' \bigr)-M_{2}\bigl(\widetilde{T}''\bigr) ={}&8+ \bigl(d_{\widetilde{T}''}(v_{k+1})+m\bigr) \biggl(m+\sum _{x\in N_{\widetilde {T}''}(v_{k+1})}d_{\widetilde{T}''}(x)\biggr) \\ &-(m+2) (m+4)-d_{\widetilde{T}''}(v_{k+1})\sum _{x\in N_{\widetilde {T}''}(v_{k+1})}d_{\widetilde{T}''}(x) \\ ={}&m\biggl[d_{\widetilde{T}''}(v_{k+1})+\sum_{x\in N_{\widetilde {T}''}(v_{k+1})}d_{\widetilde{T}''}(x)-6 \biggr] \\ \geq{}&0 .\end{aligned} $$
Also, \(M_{2}(T''')-M_{2}(\widetilde{T}'')=0\) if and only if at least one of the following conditions holds:
  1. (1)

    \(m=0\), namely, \(\widetilde{T}''\cong T'''\);

     
  2. (2)

    \(d_{\widetilde{T}''}(v_{k})=d_{\widetilde {T}''}(v_{k+1})=d_{\widetilde{T}''}(v_{k+2})=2\).

     
If \(i_{1}\neq1\) and \(i_{1}=k\), then
$$\begin{aligned} M_{2}\bigl(T''' \bigr)-M_{2}\bigl(\widetilde{T}''\bigr) ={}&4+ \bigl(d_{\widetilde{T}''}(v_{k+1})+m\bigr) \biggl(m+2+\sum _{x\in N_{\widetilde {T}''}(v_{k+1})\setminus\{v_{k}\}}d_{\widetilde{T}''}(x)\biggr) \\ &-(m+2) (m+2)-d_{\widetilde{T}''}(v_{k+1}) \biggl(\sum _{x\in N_{\widetilde {T}''}(v_{k+1})\setminus\{v_{k}\}}d_{\widetilde{T}''}(x)+m+2\biggr) \\ ={}&m\biggl(\sum_{x\in N_{\widetilde{T}''}(v_{k+1})\setminus\{v_{k}\} }d_{\widetilde{T}''}(x)-2\biggr) \\ \geq{}&0 .\end{aligned} $$
As a result, \(M_{2}(T''')-M_{2}(\widetilde{T}'')=0\) if and only if at least one of the following conditions holds:
  1. (1)

    \(m=0\), which implies that \(\widetilde{T}''\cong T'''\);

     
  2. (2)

    \(d_{\widetilde{T}''}(v_{k+1})=d_{\widetilde{T}''}(v_{k+2})=2\).

     

In what follows, we prove \(M_{1}(T''')\leq(n-k\gamma_{k})(n-k\gamma _{k}+1)+4(k\gamma_{k}-1)\) and \(M_{2}(T''')\leq(n-k\gamma_{k})[n-(k-1)\gamma _{k}]+(4k-2)\gamma_{k}-4\) 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 \(n\geq(k+1)\gamma_{k}\) by Lemma 3.3. Assume that \(\gamma _{k}\geq4\), the statement holds for \(\gamma_{k}-1\) and all the \(n\geq (k+1)(\gamma_{k}-1)\).

In view of \({\gamma_{k}(T'''-\{v_{0},v_{1},\ldots,v_{k}\})}=\gamma _{k}-1\) and \({|V(T'''-\{v_{0},v_{1},\ldots,v_{k}\})|}=n-k-1\geq (k+1)(\gamma_{k}-1)\), by the induction hypothesis, we get
$$\begin{aligned} M_{1}\bigl(T'''\bigr)&= M_{1}\bigl(T'''- \{v_{0},v_{1},\ldots,v_{k}\} \bigr)+2d_{T'''}(v_{k+1})-1+ \sum_{i=0}^{k}d^{2}_{T'''}(v_{i}) \\ &\leq M_{1}(T_{n-k-1,k,\gamma_{k}-1})+2(n-k\gamma_{k})+4k \\ &=(n-k\gamma_{k}) (n-k\gamma_{k}+1)+4(k \gamma_{k}-1).\end{aligned} $$
The equality holds if and only if \(T'''-\{v_{0},v_{1},\ldots,v_{k}\}\cong T_{n-k-1,k,\gamma_{k}-1}\) and \(d_{T'''}(v_{k+1})=\triangle=n-k\gamma_{k}\). Recalling that \(d_{T'''}(v_{i})=2\) for \(i=1,\ldots,k\), 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, \(M_{1}(T)\leq M_{1}(T')\leq M_{1}(T'')\leq M_{1}(T''')\leq(n-k\gamma _{k})(n-k\gamma_{k}+1)+4(k\gamma_{k}-1)\) 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. (1)

    \(T\cong T'\cong T''\cong T'''\cong T_{n,k,\gamma_{k}}\);

     
  2. (2)

    \(T\cong T'\cong T''\), 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
$$n-(k+1)\gamma_{k}\geq\big|N_{T''}(v_{i_{0}})\setminus \{v_{i_{0}-1},v_{i_{0}+1}\}\big|. $$
Hence,
$$\begin{aligned} d_{T''}(v_{k+1})&=d_{T'''}(v_{k+1})-|N_{T''}(v_{i_{0}}) \setminus\{ v_{i_{0}-1},v_{i_{0}+1}\}| \\ &\geq d_{T'''}(v_{k+1})-\bigl[n-(k+1)\gamma_{k}\bigr] \\ &=\gamma_{k}\geq3,\end{aligned} $$
a contradiction to \(d_{T''}(v_{k+1})=2\). Therefore,
$$M_{1}(T)\leq(n-k\gamma_{k}) (n-k\gamma_{k}+1)+4(k \gamma_{k}-1) $$
with equality if and only if \(T\cong T_{n,k,\gamma_{k}}\).
Note that \({\gamma_{k}(T'''-\{v_{0},\ldots,v_{k-1}\})}=\gamma _{k}-1\) and \({|V(T'''-\{v_{0},\ldots,v_{k-1}\})|}>(k+1)(\gamma _{k}-1)\). Then
$$\begin{aligned} M_{2}\bigl(T'''\bigr)&= M_{2}\bigl(T'''- \{v_{0},v_{1},\ldots,v_{k-1}\} \bigr)+d_{T'''}(v_{k+1})+4(k-1)+2 \\ &\leq M_{2}(T_{n-k,k,\gamma_{k}-1})+n-k\gamma_{k}+4(k-1)+2 \\ &=(n-k\gamma_{k})\bigl[n-(k-1)\gamma_{k}\bigr]+(4k-2) \gamma_{k}-4.\end{aligned} $$
The equality holds if and only if \(T'''-\{v_{0},\ldots,v_{k-1}\}\cong T_{n-k,k,\gamma_{k}-1}\) and \(d_{T'''}(v_{k+1})=\triangle=n-k\gamma _{k}\). In consideration of \(d_{T'''}(v_{i})=2\) for \(i=1,\ldots,k\), the equality holds if and only if \(T'''\cong T_{n,k,\gamma_{k}}\).
Hence, if \(i_{1}\neq1\), then \(M_{2}(T)\leq M_{2}(T')\leq M_{2}(\widetilde {T}'')\leq M_{2}(T''')\leq(n-k\gamma_{k})[n-(k-1)\gamma_{k}]+(4k-2)\gamma _{k}-4\), with equality if and only if at least one of the following conditions holds:
  1. (1)

    \(T\cong T'\cong\widetilde{T}''\cong T'''\cong T_{n,k,\gamma_{k}}\);

     
  2. (2)

    \(T\cong T'\cong\widetilde{T}''\), where \(d_{\widetilde {T}''}(v_{k})=d_{\widetilde{T}''}(v_{k+1})=d_{\widetilde {T}''}(v_{k+2})=2\) and \(\widetilde{T}'''\cong T_{n,k,\gamma_{k}}\).

     
Analogous to the analysis of the first Zagreb index, the second condition above is impossible. Thus,
$$M_{2}(T)\leq(n-k\gamma_{k})\bigl[n-(k-1)\gamma_{k} \bigr]+(4k-2)\gamma_{k}-4 $$
and the equality holds if and only if \(T\cong T_{n,k,\gamma_{k}}\).

Besides, if \(i=1\), then \(M_{2}(T)\leq(n-k\gamma_{k})[n-(k-1)\gamma _{k}]+(4k-2)\gamma_{k}-4\) 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
$$M_{1}(T)\leq(n-\gamma) (n-\gamma+1)+4(\gamma-1) $$
and
$$M_{2}(T)\leq 2(n-\gamma+1) (\gamma-1)+(n-\gamma) (n-2\gamma+1), $$
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 \(k\geq2\). Note that the domination number is the special case of the distance k-domination number for \(k=1\) and \(T_{n,k,\gamma_{k}}\cong T_{n,\gamma}\), \(T_{n,k,2}^{i}\cong T_{n,\gamma}\), \(i\in\{1,\ldots,k\}\), 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
$$M_{1}(T)\leq \textstyle\begin{cases}n(n-1) & \text{if }\gamma _{k}=1,\\ (n-k\gamma_{k})(n-k\gamma_{k}+1)+4(k\gamma_{k}-1) &\text{if }\gamma_{k}\geq2, \end{cases} $$
with equality if and only if \(T\cong S_{n}\) when \(\gamma_{k}=1\), \(T\cong T_{n,k,2}^{i}\), \(i\in\{1,\ldots,k\}\), when \(\gamma_{k}=2\), or \(T\cong T_{n,k,\gamma_{k}}\) when \(\gamma_{k}\geq3\). Moreover,
$$M_{2}(T)\leq \textstyle\begin{cases}2(n-\gamma_{k}+1)(\gamma_{k}-1)+(n-\gamma_{k})(n-2\gamma _{k}+1) & \text{if }k=1,\\ (n-1)^{2} & \text{if }k\geq2, \gamma_{k}=1,\\ (n-k\gamma_{k})[n-(k-1)\gamma_{k}]+(4k-2)\gamma _{k}-4 & \text{if }k\geq2, \gamma_{k}\geq2, \end{cases} $$
with equality if and only if \(T\cong S_{n}\) when \(k\geq2\) and \(\gamma _{k}=1\), \(T\cong T_{n,k,2}^{i}\), \(i\in\{2,\ldots,k\}\), when \(k\geq2\) and \(\gamma_{k}=2\), or \(T\cong T_{n,k,\gamma_{k}}\) otherwise.

Declarations

Acknowledgements

This work is financially supported by the National Natural Science Foundation of China (No. 11401004) and the Natural Science Foundation of Anhui Province of China (No. 1408085QA03).

Authors’ contributions

All authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Mathematical Sciences, Anhui University, Hefei, Anhui, 230601, China

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