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

     这道题也挺有意思的。不过首先要搞懂题目，首先搞清楚怎样的BTS才算symmetric的。必须要是一个balanced的BTS然后两边的height都相等。

     这道题个人觉得用recursive来写比较容易哈。

     代码如下。~

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean isSymmetric(TreeNode root) {        if(root==null) return true;        return symhelper(root.left,root.right);    }        public boolean symhelper(TreeNode left,TreeNode right){        if(left==null||right==null){            return (left==right);        }        if((left.val==right.val)&&(symhelper(left.left,right.right))&&(symhelper(left.right,right.left))){            return true;        }else{            return false;        }    }}