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;
}
}
}