- 分类：Tree/BackTracking
- 时间复杂度: O(n) 这种把树的节点都遍历一遍的情况时间复杂度为O(n)
- 空间复杂度: O(h) 树的节点的深度
101. Symmetric Tree
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
Note:
Bonus points if you could solve it both recursively and iteratively.
代码：
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isSymmetric(self, root: 'TreeNode') -> 'bool':
if root==None:
return True
res=self.helper(root.left,root.right)
return res
def helper(self,left,right):
if left==None and right==None:
return True
elif left==None or right==None:
return False
else:
if left.val==right.val:
return self.helper(left.right,right.left) and self.helper(left.left,right.right)
else:
return False
讨论：
1.这道题是道简单的好题，思路看代码就成了！
