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 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) and 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)
##### 递归，判断是否对称