Spanning Trees In K_1,5-free Graphs

Let G,H be simple graph, if there is no induced subgraph in G is isomorphic to H.then we call G H-free graph. L. Gargano and his team implied that:If a K1,3-free graph G, satisfies σk+3(G)≥n-k-2, then G admits a spanning tree with at most k branching vertices. E. Flandrin and his team pose the conjecture:Every connected K1,4-free graph G with σ4(G)≥|G|contains a spanning tree with at most3leaves, and this conjecture was solved by A. Kyaw In2009.In this paper, we use reduction ad absurdum to study the structure and properties of the tree with5leaves.and prove that when the degree of independent number meets certain conditions, there is a spanning tree with at most4leaves in G, the specific conclusions is as follows:Let G be a connected K1,5-free graph, If σ5(G)≥|G|+18, then G has a spanning tree with at most4leaves.