每日一道Leetcode - 剑指 Offer 28. 对称的二叉树【递归】

# 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:
        def recur(L,R):
            # 如果左右树都为空，则为遍历完的情况，返回True
            if not L and not R:
                return True
            # 一个为空一个不为空或者当前的值不同，返回False
            if not L or not R or L.val!=R.val:
                return False
            # 判断左子树的左边和右子树的右边是否相同，均相同才返回True
            return recur(L.left,R.right) and recur(L.right,R.left)
    
        # 判断root是否为空，空则返回True,非空则递归判断子树情况
        return recur(root.left,root.right) if root else True

标签：right,return,recur,Offer,root,self,28,二叉树,True
来源： https://blog.csdn.net/weixin_41041275/article/details/114738233