26-树的子结构

题目

输入两棵二叉树A和B，判断B是不是A的子结构。(约定空树不是任意一个树的子结构)
B是A的子结构， 即 A中有出现和B相同的结构和节点值。
https://leetcode-cn.com/problems/shu-de-zi-jie-gou-lcof/

思路

1. 在树A中找到和树B的根节点的值一样的节点
2. 判断树A中以R为节点的子树是不是包含和树B一样的结构

代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def isSubStructure(self, A: TreeNode, B: TreeNode) -> bool:
        def DoesTree1HaveTree2(A, B):
            if B is None:
                # 说明遍历完B都没有return False退出
                return True
            if A is None:
                return False
            if A.val != B.val:
                return False
            return DoesTree1HaveTree2(A.left, B.left) and DoesTree1HaveTree2(A.right, B.right)
        result = False
        if A != None and B != None:
            if A.val == B.val:
                result = DoesTree1HaveTree2(A, B)
            if not result:
                result = self.isSubStructure(A.left, B)
            if not result:
                result = self.isSubStructure(A.right, B)
        return result