题解 | #对称的二叉树#

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param pRoot TreeNode类 
# @return bool布尔型
#
from collections import deque

class Solution:
    def isSymmetrical(self , pRoot: TreeNode) -> bool:
        # write code here
        if not pRoot:
            return True
        queue = deque([pRoot.left, pRoot.right])
        while queue:
            size = len(queue)
            if size % 2:
                return False
            level_values = []
            for _ in range(size):
                node = queue.popleft()
                if node:
                    level_values.append(node.val)
                    queue.append(node.left)
                    queue.append(node.right)
                else:
                    level_values.append(None)
            if level_values != level_values[::-1]:
                return False
        return True

使用层次遍历直接对比每层的结果即可。