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
#
# 主要就是判断形状，另外当进入空节点时，要对其进行判断
class Solution:
    def isSymmetrical(self, pRoot: TreeNode) -> bool:
        if not pRoot or (not pRoot.left and not pRoot.right):
            return True

        def mir(left, right):
            type = 0

            try:
                if not right.right and left.left:
                    return False
            except:
                type += 1
            try:
                if not right.left and left.right:
                    return False
            except:
                type += 2
            try:
                if not left.left and right.right:
                    return False
            except:
                type += 4
            try:
                if not left.right and right.left:
                    return False
            except:
                type += 8

            if not left and not right:
                return True
            if left.val != right.val:
                return False
            if (
                (not left.left)
                and (not left.right)
                and (not right.right)
                and (not right.left)
            ):
                return True

            if type == 0:
                return mir(left.left, right.right) and mir(left.right, right.left)
            elif type == 1:
                return mir()

        if (pRoot.left and not pRoot.right) or (not pRoot.left and pRoot.right):
            return False

        return mir(pRoot.left, pRoot.right)