题解 | #二叉树的镜像#

# 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
        lStack = []
        rStack = []
        lStack = self.qBianli(pRoot.left, lStack)
        rStack = self.hBianli(pRoot.right, rStack)
        if lStack == rStack:
            return True
        else:
            return False
        
    # 前序遍历 左
    def qBianli(self, pRoot: TreeNode, lStack:list):
        if pRoot:
            self.qBianli(pRoot.left, lStack)
            self.qBianli(pRoot.right, lStack)
            lStack.append(pRoot.val)
        else:
            lStack.append(0)
        return lStack
        
    def hBianli(self, pRoot: TreeNode, rStack:list):
        if pRoot:
            self.hBianli(pRoot.right, rStack)
            self.hBianli(pRoot.left, rStack)
            rStack.append(pRoot.val)
        else:
            rStack.append(0)
        return rStack