剑指Offer 17. 树的子结构 （二叉树）

题目描述

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

https://www.nowcoder.com/practice/6e196c44c7004d15b1610b9afca8bd88?tpId=13&tqId=11170&rp=3&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking

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)

原文地址：https://www.cnblogs.com/huangqiancun/p/9784299.html