标签：and amp type binary this solution app check div
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1 / 2 2 / \ / 3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1 / 2 2 \ 3 3
BFS and Iterative:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def isSymmetric(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
que=[root]
while que:
check=[]
n=len(que)
for i in range(n):
node=que.pop(0)
if node:
que.append(node.left)
que.append(node.right)
check.append(node.val)
else:
check.append(None)
n=len(check)
for i in range(n):
if check[i]!=check[n-i-1]:
return False
return True
Recursion:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def isSymmetric(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
def findsys(node1,node2):
if node1==None and node2==None:
return True
if node1==None or node2==None:
return False
return node1.val==node2.val and findsys(node1.left,node2.right) and findsys(node1.right,node2.left)
return findsys(root,root)
[LeetCode&Python] Problem 101. Symmetric Tree
标签：and amp type binary this solution app check div
原文地址：https://www.cnblogs.com/chiyeung/p/10200475.html
