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 isSym_lr(self,lroot: TreeNode, rroot: TreeNode) -> bool: # 传入左右两个子树，看是否对称
        issym = False
        if lroot.val == rroot.val: # 想变成ture 前提val相等
            if lroot.left and rroot.right and not(lroot.right or rroot.left):  # 传入左树的左边和右树的右边
                issym = self.isSym_lr(lroot.left, rroot.right)  # 看子树是否对称
            if lroot.right and rroot.left and not(lroot.left or rroot.right):  # 看另一边
                issym = self.isSym_lr(lroot.right, rroot.left)
            if lroot.left and rroot.right and lroot.right and rroot.left:
                issym = self.isSym_lr(lroot.right, rroot.left) and self.isSym_lr(lroot.left, rroot.right)


            if not (lroot.right or lroot.left or rroot.left or rroot.right):
                # 如果两边都没有子节点并且val相等，则True
                issym = True
        return issym

    def isSymmetrical(self,pRoot: TreeNode):
        if not pRoot or (not pRoot.left and not pRoot.right):
            return True
        if not pRoot.left or not pRoot.right:
            return False
        return self.isSym_lr(pRoot.left, pRoot.right)