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

這道題也挺有意思的。不過首先要搞懂題目，首先搞清楚怎樣的BTS才算symmetric的。必須要是一個balanced的BTS然后兩邊的height都相等。

這道題個人覺得用recursive來寫比較容易哈。

代碼如下。~

/** * 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;        }    }}