[leetcode tree]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

判断一棵树是否对称

递归:

 1 class Solution(object):
 2     def isSymmetric(self, root):
 3         if not root:
 4             return True
 5         return self.issym(root.left,root.right)
 6     def issym(self,left,right):
 7         if not left and not right:
 8             return True
 9         if not left or not right:
10             return False
11         if left.val == right.val:
12             return self.issym(left.left,right.right) and self.issym(left.right,right.left)
13         return False

非递归:

递归本质上就是栈,所以。。。

 1 class Solution(object):
 2     def isSymmetric(self, root):
 3         if not root:
 4             return True
 5         stack=[[root.left,root.right]]
 6         while stack:
 7             left,right = stack.pop()
 8             if not left and not right:
 9                 continue
10             if not left or not right:
11                 return False
12             if left.val == right.val:
13                 stack.append([left.right,right.left])
14                 stack.append([left.left,right.right])
15             else:
16                 return False
17         return True
原文地址:https://www.cnblogs.com/fcyworld/p/6522751.html