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
# 递归判断是否镜像
class Solution:
    def isSymmetrical(self , pRoot: TreeNode) -> bool:
        # write code here
        return self.tojudge(pRoot, pRoot)

    def tojudge(self, p , q): # p，q为待测的镜像两个节点
        if not p and not q:
            return True # 都空，不用继续判断了，返回真
        if p and not q:
            return False
        if q and not p:
            return False # 不对称则假
        if p and q:
            if p.val != q.val: # 对称不相等也假
                return False
            else: # 对称且相等则递归下去判断其子树对不对称
                return self.tojudge(p.left, q.right) and self.tojudge(p.right, q.left)