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
#
class Solution:
    def isSymmetrical(self , pRoot: TreeNode) -> bool:
        # write code here
        if not pRoot:
            return True
        q = []
        if pRoot.left:
            q.append(pRoot.left)
        if pRoot.right:
            q.append(pRoot.right)
        
        while q:
            if len(q) % 2 != 0:
                return False
            n = len(q)
            left = []
            right = []
            left_node = []
            right_node = []
            for _ in range(n//2):
                node_l = q.pop(0)
                if node_l:
                    left.append(node_l.val)
                    left_node.append(node_l)
                else:
                    left.append(node_l)
            for _ in range(n // 2):
                node_r = q.pop()
                if node_r:
                    right_node.append(node_r)
                    right.append(node_r.val)
                else:
                    right.append(node_r)
            if left == right:
                for node in left_node:
                    q.append(node.left)
                    q.append(node.right)
                for node in right_node[::-1]:
                    q.append(node.left)
                    q.append(node.right)
            else:
                return False
        return True