101.对称二叉树

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        if root is None:
            return True
        
        return self.isSymmetricRecu(root.left, root.right)
    
    def isSymmetricRecu(self, left, right):
        if left is None and right is None:
            return True
        if left is None or right is None or left.val != right.val:
            return False
        return self.isSymmetricRecu(left.left, right.right) and self.isSymmetricRecu(left.right, right.left)

 

posted @ 2019-09-03 19:14  我叫郑小白  阅读(106)  评论(0编辑  收藏  举报