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 recursion(self,root1, root2) -> bool:
        # 两个都为空 直接返回ture
        if not root1 and not root2:
            return True
        # 判断为空必须放前面 隔离 None没有val的错误
        if root1 == None or root2 == None or root1.val != root2.val:
            return False
        return self.recursion(root1.left, root2.right) and self.recursion(root1.right, root2.left) 
    

    def isSymmetrical(self , pRoot: TreeNode) -> bool:
        if not pRoot:
            return True
        return self.recursion(pRoot.left, pRoot.right)