题解 | #对称的二叉树#

对称的二叉树

描述 给定一棵二叉树，判断其是否是自身的镜像（即：是否对称）

数据范围：节点数满足 0≤n≤1000，节点上的值满足∣val∣≤1000 要求：空间复杂度 O(n)，时间复杂度 O(n) 备注： 你可以用递归和迭代两种方法解决这个问题

思路：题目挺简单的，将pRoot分成左子树leftTree和右子树rightTree，leftTree的每一个子树的左边节点left的值等于rightTree的每一个子树的右边节点right的值，leftTree的每一个的右边节点right的值等于rightTree的左边节点left的值，递归一下就行了。

# 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 isBanceTree(self,leftTree,rightTree):
        if (leftTree == None) & (rightTree == None):
            return True
        elif (leftTree != None) & (rightTree == None):
            return False
        elif (leftTree == None) & (rightTree != None):
            return False
        elif leftTree.val != rightTree.val:
            return False
        else:
            return self.isBanceTree(leftTree.left, rightTree.right) & self.isBanceTree(leftTree.right, rightTree.left)
    def isSymmetrical(self , pRoot: TreeNode) -> bool:
        # write code here
        if pRoot == None:
            return True
        return self.isBanceTree(pRoot.left, pRoot.right)