输入两棵二叉树A，B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构）
题目地址
思路
如果树1或树2为空时，返回False
在树1中找到和树2一样的根结点R，然后在判断树1中以R为根结点的子树是否与树2有一样的结构。
先以树1 的根结点为起点寻找是否包含树B，如果找不到就以树A的左结点为起点寻找，若还找不到，以树A的右结点为起点寻找，递归进行
Python
# -*- coding:utf-8 -*- class TreeNode: def __init__(self,x): self.val = x self.left = None self.right = None node1 = TreeNode(1) node2 = TreeNode(2) node3 = TreeNode(3) node4 = TreeNode(4) node5 = TreeNode(5) node1.left = node2 node2.right = node3 node3.left = node4 node3.right = node5 class Solution: def HasSubtree(self, pRoot1, pRoot2): # write code here if not pRoot1 or not pRoot2: return False result = False if pRoot1.val == pRoot2.val: # 找到根结点相同的结点 result = self.isSubTree(pRoot1, pRoot2) if not result: # 没找到，在左子树寻找 result = self.HasSubtree(pRoot1.left, pRoot2) if not result: # 还没找到，在右子树寻找 result = self.HasSubtree(pRoot1.right, pRoot2) return result def isSubTree(self,tree1, tree2): if not tree2: return True if not tree1: return False if tree1.val != tree2.val: return False return self.isSubTree(tree1.left,tree2.left) and self.isSubTree(tree1.right,tree2.right) if __name__ == ‘__main__‘: result = Solution().HasSubtree(node1, node3) print(result)
原文：https://www.cnblogs.com/huangqiancun/p/9784299.html