问题描述：

 

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.

 

思路：二分类树的问题，可以考虑递归解法

 

# Definition for a binary tree node.# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution:    def isSymmetric(self, root: TreeNode) -> bool:        if root == None:            return True        def ismirror(root1,root2):            if root1 == None and root2 ==None:                return True            elif root1 == None or root2 ==None or root1.val != root2.val:                return False            return ismirror(root1.left,root2.right) and ismirror(root1.right,root2.left)                 return ismirror(root.left,root.right)