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Symmetric Tree

时间:2016-12-22 07:15:18      阅读:200      评论:0      收藏:0      [点我收藏+]

标签:nod   this   null   init   blog   self   root   div   round   

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

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public boolean isSymmetric(TreeNode root) {
12         if (root == null) return true;
13         return helper(root.left, root.right);
14     }
15     
16     private boolean helper(TreeNode left, TreeNode right) {
17         if (left == null && right == null) return true;
18         
19         if (left == null || right == null ) return false;
20         
21         if (left.val != right.val) return false;
22         
23         return helper(left.left, right.right) && helper(left.right, right.left);
24     }
25 }

 

Symmetric Tree

标签:nod   this   null   init   blog   self   root   div   round   

原文地址:http://www.cnblogs.com/beiyeqingteng/p/6209753.html

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