这里A这道题的时候，刚看到题目的时候想到用递归。后面题目也有提示，用递归或迭代。还是没有弄出来，主要参考了

LeetCode Oj:

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

1   / \  2   2 / \ / \3  4 4  3

 

But the following is not:

1   / \  2   2   \   \   3    3

 

Note:

Bonus points if you could solve it both recursively and iteratively.

confused what "{1,#,2,3}" means? 

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

1  / \ 2   3    /   4    \     5

The above binary tree is serialized as 

"{1,2,3,#,#,4,#,#,5}".

 

 

1 class TreeNode { 2      int val; 3      TreeNode left; 4      TreeNode right; 5      TreeNode(int x) { val = x; } 6   } 7  8 public class Solution { 9     public boolean isSymmetric(TreeNode root) {        10         if(null == root)11             return true;12         return check(root.left, root.right);13     }14     public boolean check(TreeNode leftNode, TreeNode rightNode){
    //开始有想到这样，用左右子节点15         if(null == leftNode && null == rightNode) //两个都是空节点16             return true;17         if(null == leftNode || null == rightNode)//其中一个空节点18             return false;19         //两个都不是空节点20         return leftNode.val == rightNode.val && check(leftNode.left, rightNode.right) 21                 && check(leftNode.right, rightNode.left);//这里用的很漂亮，主要逻辑也在这里22     }23 }