#### Abstract

A graph *G* is called -free if *G* contains no as an induced subgraph. A tree with at most *m* leaves is called an *m*-ended tree. Let be the minimum degree sum of *k* independent vertices in *G*. In this paper, it is shown that every connected -free graph *G* contains a spanning 6-ended tree if .

#### 1. Introduction

We will begin with some basic definitions, notations, and terminologies used by Bondy and Murty [1]. The *graph* generally means a simple undirected graph *G* with a vertex set and an edge set . We write for the order of *G*. For any , let be the set of vertices adjacent to in *G* and . For a nonempty subset , we write

For an integer *i*, we define and . Let *T* be a tree. A vertex of degree one in *T* is called a leaf of *T*, and a vertex of degree at least three is called a branch vertex. If are vertices of *T*, the path in *T* connecting *u* and is unique and is denoted by . Define , and . The distance between *u* and in *T* is denoted by .

A subset is called an independent set of *G* if no two vertices of *A* are adjacent in *G*. The independence number of *G*, , is the order of the maximum independent set of *G*. For an integer *k*, we define

For an integer , a graph *G* is called -free if *G* contains no as an induced subgraph. A tree with at most *m* leaves is called an *m*-ended tree. If *T* is an *m*-ended tree in *G* such that is as large as possible, we call *T* the maximum *m*-ended tree of *G*. There are several well-known conditions ensuring that a graph *G* contains a spanning *m*-ended tree (see the survey paper [2]). In 1979, Win [3] obtained a sufficient condition related to an independent number for *k*-connected graphs, which confirms the conjecture of Las Vergnas [4].

Theorem 1 (Win [3]). *Let G be a k-connected graph, and let . If , then G has a spanning m-ended tree.*

There are many results on the degree sum conditions for a graph to contain a spanning tree with a bounded number of leaves or branch vertices. Among them are the following theorems.

Theorem 2 (Broersma and Tuinstra [5]). *Let G be a connected graph with n vertices, and let . If , then G has an m-ended tree.*

Theorem 3 (Gargano et al. [6]). *Let k be a nonnegative integer, and let G be a connected claw-free graph. If , then G has a spanning tree with at most k branch vertices.*

Theorem 4 (Kano et al. [7]). *Let k be a nonnegative integer, and let G be a connected claw-free graph. If , then G has a spanning tree with at most leaves.*

Theorem 5 (Kyaw [8]). *Every connected -free graph with contains a spanning tree with at most 3 leaves.*

Theorem 6 (Kyaw [9]). * Let G be a connected -free graph.*(1)

*If , then*(2)

*G*has a Hamiltonian path.*If for an integer , then*

*G*has a spanning tree with at most*k*leaves.Theorem 7 (Chen et al. [10]). *Let G be a connected -free graph with n vertices. If , then G contains a spanning 4-ended tree.*

In this paper, we further consider -free graphs and generalize Theorem 7 as follows.

Theorem 8. *Let G be a connected -free graph. If , then G contains a spanning 6-ended tree.*

Before proving Theorem 8, we first give an example to show that the condition in Theorem 8 is sharp.

Let *m* be a positive integer, and let *G* be the graph obtained from a triangle and 7 copies of by joining *x* to each vertex of , *y* to each vertex of , and *z* to each vertex of (see Figure 1). It is easy to see that , but each spanning tree of *G* has at least 7 leaves. Therefore, the condition in Theorem 8 is sharp.

#### 2. Proof of Theorem 8

In this section, we will prove Theorem 8. First, we prove the following two lemmas.

Lemma 1. *Let G be a connected graph with no spanning 6-ended tree, and let T be a maximum 7-ended tree of G. Then, G does not contain a 6-ended tree such that .*

*Proof of Lemma 1. *By way of contradiction, suppose that there is a 6-ended tree in *G* such that . Since *G* has no spanning 6-ended tree, is not a spanning tree. Hence, . As *G* is connected, there are two vertices such that and . Noting that has at most 7 leaves, we conclude that is a 7-ended tree, contrary to the maximality of *T*. Hence, Lemma 1 is true.

Lemma 2. *Let T be a tree with n vertices. For an integer , if T has at most k leaves, then it has at most branch vertices.*

*Proof of Lemma 2. *Assume that *T* has *x* vertices of degree two, *y* branch vertices, and *z* leaves. Then, and . Hence, . This completes the proof of Lemma 2.

Based on the above two lemmas, we can now prove the main result of this paper.

*Proof of Theorem 8. *Let *G* be a connected -free graph. Suppose, by contradiction, that *G* does not contain a spanning 6-ended tree. As *G* is connected, there is a spanning tree *T* in *G*. By our hypothesis, i.e., *G* does not contain a spanning 6-ended tree, the number of leaves in *T* is at least 7. If the number of leaves in *T* is 7, then we have found a tree with exactly 7 leaves. If the number of leaves in *T* is more than 7, then we can delete some of the leaves, and hence, there always exists a 7-ended tree in *G*. We choose a maximum 7-ended tree *T*. By Lemma 1, *T* has exactly 7 leaves. Denote by the set of leaves of *T*, and let be the set of branch vertices of *T*. From Lemma 2, we can conclude that . By the maximality of *T*, we have . For , let be the unique component of containing , and let be the only neighbor of *S* in . Denote by the only neighbor of in *S*. For convenience, we let . Note that . For , let be the predecessor of *x* and be the successor of *x*. If , let be the longest path between the branch vertices. Without loss of generality, we may assume that and . For convenience, denote by the predecessor of and the successor of for all . Set . If for some and , then we call the branch vertex for .

*Claim 1. *For all with , if , then .

*Proof of Claim 1. *By way of contradiction, assume Claim 1 is false. SetThen, is a 6-ended tree with , which is contrary to Lemma 1. This completes the proof of Claim 1.

It follows from Claim 1 that *U* is an independent set. Since *G* is -free, .

*Claim 2. *For , .

*Proof of Claim 2. *By way of contradiction, assume that there exists a vertex for some . Then, there exist three distinct integers such that . By Claim 1, we can derive that and . As , . Set . Then, is a 6-ended tree with , which is a contradiction. This completes the proof of Claim 2.

By an argument similar to the proof of Claim 2, we can derive that , .

*Claim 3. * for all *.*

*Proof of Claim 3. *Suppose, on the contrary, that there are two vertices for some . Without loss of generality, assume that . By Claim 1, . Assume that for some . Since , there exists such that .

If , then . Then, is a 6-ended tree with a vertex set , which contradicts Lemma 1. Hence, . Since *G* is -free, it can be derived that . Set . Then, is a 6-ended tree with , which is contrary to Lemma 1. This completes the proof of Claim 3.

*Claim 4. *If , then for all .

*Proof of Claim 4. *By way of contradiction, assume that Claim 4 is false. Then, there is a vertex *x* such that for some with . By Claim 1, . Set . Then, is a 6-ended tree with , which contradicts Lemma 1. This completes the proof of Claim 4.

*Claim 5. *For all , , , , and are pairwise disjoint, where .

*Proof of Claim 5. *It can be easily concluded that . It follows from Claim 1 that . This completes the proof of Claim 5.

By Claim 5, we haveThis together with Claim 4 implies that

*Claim 6. *If for some and , then and for some .

*Proof of Claim 6. *By way of contradiction, assume that or . SetThen, is a 6-ended tree with , which is contrary to Lemma 1. This completes the proof of Claim 6.

For every vertex , defineFor every vertex set , defineIf , we consider the following two cases.

*Case 1. *.

In this case, . We choose the maximum *T* with 7 leaves such that(C1) is as large as possible.(C2) is as small as possible, subject to (C1).Without loss of generality, we may assume that , , and . Hence, and . SetLet be the family of graphs shown in Figure 2. Then, in Case 1, .

The following claim plays a crucial role in the proof of Case 1.

*Claim 7. *Let be two consecutive vertices in such that , where and . Let be the branch vertex for , where . If *G* has no 6-ended tree, then and .

*Proof of Claim 7. *By way of contradiction, assume that . Set . Then, is a 6-ended tree with , which is contrary to Lemma 1. Similarly, . This completes the proof of Claim 7.

*Claim 8. *Let be the branch vertex for for some and . Then, . Specifically, for all and for all .

*Proof of Claim 8. *Suppose, for contradiction, that for some . Set . Hence, is a 6-ended tree with , contradicting Lemma 1. Hence, . Likewise, . This completes the proof of Claim 8.

*Claim 9. *If for some , then .

*Proof of Claim 9. *By way of contradiction, suppose and . Hence, there exist two integers with such that . By Claim 8, we can see that is not the branch vertex for or . This together with , by applying Claim 7, implies that . This and Claim 8 imply that , which is a contradiction. Hence, , and similarly, if , then .

The proof of Case 1 has been divided into 6 subcases.

*Subcase 1. *.

Without loss of generality, we may assume that , .

We choose *T* such that(C3) is as large as possible, subject to (C1) and (C2).

*Subclaim 1. *For every vertex .

*Proof of Subclaim 1. *By way of contradiction, suppose that there is a vertex such that . Then, there exist three distinct integers such that . By (C2), we can see that . If , set . Hence, is a 6-ended tree with , which is contrary to Lemma 1. Thus, . Since *G* is -free, there exists such that or , contradicting Claim 7. This completes the proof of Subclaim 1.

Set . If for every vertex , then . Otherwise, suppose that there are *s* vertices in satisfying for all . Define , , and . Therefore, .

For every , . It follows from Claim 7 and (C2) that . Hence, . Since , . Thus, . Therefore,

*Subclaim 2. *

*Proof of Subclaim 2. *If , by (C2), , say . Then, . Otherwise, assume that and for some . Then by (C2), , that is, . If , the result holds. Hence, we assume that . Since , by Claims 4 and 6, for some . SetThen, is a 6-ended tree with , which contradicts Lemma 1. It follows from (5) that

*Subclaim 3. *.

*Proof of Subclaim 3. *If there are three distinct integers in such that , then since *G* is -free, then , which contradicts (C2).

If for , by (C2), . Without loss of generality, set . Since , by (C2) and Claims 1 and 6, we have . Also by Claim 6, . Moreover, for all .

If for some , set . Then by Claim 7, . If there is a vertex such that , then . Otherwise, for every , . By Claim 7, . Specifically, . By applying Claim 7, we get a contradiction. Hence, . If for some , then . Suppose, for contradiction, and . Then by (C3), . Set , which contradicts (C2). This together with (5) implies that the result holds.

It follows from Subclaims 2-3 thatwhich implies that , which is a contradiction. This completes the proof of Subcase 1.

*Claim 10. *If and there is a vertex , then .

*Proof of Claim 10. *Assume that there is a vertex such that . Then, there exist three distinct integers such that . Set . Then, is a maximum tree with 7 leaves. Since are branch vertices, by (C2), and . This implies that we can find a tree such that . This completes the proof of Claim 10.

Thus, in the following subcases, we suppose that, for every vertex , .

If for every vertex , then . Otherwise, suppose that there are *s* vertices in *P* satisfying for all . Define , , and . Therefore, .

It follows from Claim 7, Claim 8, and (C2) that . Since , by applying Claim 7 with and (C2), we can derive that either or there exists some such that for all . In the latter case, combining with , we can conclude that contradict Claim 7. Thus, , and hence, , where .

If , similar arguments apply to and then . Thus,If , by Claim 7, we can deduce that for all . Hence,In the following, if , we choose *T* such that(C4) is as large as possible, subject to (C1) and (C2).

*Subcase 2. *.

Without loss of generality, we may assume that and . If , by (C2), .

Since , there are three integers in , say , such that or for some . Hence, and for all .â€‰(i) . In this case, and for all . If , by Claim 6 and (C2), we can derive that . Hence, .â€‰(ii) . Then, , since otherwise is a 6-ended tree with , which contradicts Lemma 1. If , then assume that and for some . By (C2), . Setâ€‰Then, is a 6-ended tree with , which contradicts Lemma 1. Thus, . Hence, .(iii). Then by similar arguments, . If , then we claim that . If , then the above claim is true. Hence, . Assume and . By similar arguments, we have . Thus, . Since , by Claims 1, 4, and 6, , where . SetThen, is a 6-ended tree with , which contradicts Lemma 1. Hence, . Above all, we haveIf there are three distinct integers such that , then . Since , we have for some , which contradicts Claim 7.

From (C4), for all . If , then .

If , there is an integer such that . By symmetry, set . Then, and . Otherwise, if , set . If , then . Set . Then, is a 6-ended tree with , which contradicts Lemma 1. Moreover, . This together with the proof of (14) implies that .

In either case, we can getwhich implies that , which is a contradiction. This completes the proof of Subcase 2.

*Subcase 3. *.

In this case, *T* satisfies and . Without loss of generality, we may assume that and .

If , then . Otherwise, we can find a tree satisfying Subcase 2. Analysis similar to that of Subcase 2 shows thatIf for some , then set . Then, . If , the result is true. Hence, . Suppose, for contradiction, for some . By (C4), we have . Since *G* is -free, for some . SetThen, is a 6-ended tree with , which contradicts Lemma 1. Hence, if for some , then .

If for some , then set . Then, and , that is, .

If for some , then set . Since , for some , say . Hence, . By similar arguments, we can derive that if , then . Hence,Above all, we haveThis together with (20) shows thatwhich implies that , which is a contradiction. This completes the proof of Subcase 3.

*Subcase 4. *.

In this case, *T* satisfies and . Without loss of generality, we may assume that , .

*Subclaim 4. * for all . Similarly, for all and for all .

*Proof of Subclaim 4. *By way of contradiction, suppose that for some . Set . Then, , which is a contradiction to Subcase 2. Similarly, suppose that or for some . Set or . Then, , which is a contradiction to Subcase 3. This completes the proof of Subclaim 4.

*Subclaim 5. *.

*Proof of Subclaim 5. *It follows from (C4) thatIt , then . By way of contradiction, assume that and . By (25), . Set . Then, , which contradicts (C2).

If for , then . Hence, . By Claim 7, . Moreover, . Otherwise, assume that . Then, . Set . is a 6-ended tree with , which is contrary to Lemma 1. Hence, . By a similar argument as in the proof of (10), we haveThus, . This completes the proof of Subclaim 5.

*Subclaim 6. *.

*Proof of Subclaim 6. *From (C4), we haveIf such that , then we assert that . By way of contradiction, assume that and