题解 | #对称的二叉树#
对称的二叉树
http://www.nowcoder.com/practice/ff05d44dfdb04e1d83bdbdab320efbcb
```# 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
不知道是不是不想要我了，求阿里收留
然后现在约到了字节的一面和淘天的复活赛，外加猿辅导。华为笔试完没动静。
美团那边之前投了个base广州的，把我流程卡麻了，应该是不怎么招人，我直接简历挂了，现在进了一个正常的后端流程，还在筛选，不知道还有没有hc。