码迷,mamicode.com
首页 > 其他好文 > 详细

Binary Tree Upside Down

时间:2016-11-15 07:37:25      阅读:104      评论:0      收藏:0      [点我收藏+]

标签:efi   方法   example   init   and   分析   root   实现   nod   

 Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

For example:
Given a binary tree {1,2,3,4,5},

    1
   /   2   3
 / 4   5

 

return the root of the binary tree [4,5,2,#,#,3,1].

   4
  /  5   2
    /    3   1  

 分析:

对于树,递归总是是很好的实现方法,所以这题应该

 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 TreeNode upsideDownBinaryTree(TreeNode root) {
12         if (root == null || root.left == null) {
13             return root;
14         }
15         TreeNode newRoot = upsideDownBinaryTree(root.left);
16         
17         root.left.right = root;
18         root.left.left = root.right;
19         
20         root.left = null;
21         root.right = null;
22         
23         return newRoot;
24     }
25 }

 

Binary Tree Upside Down

标签:efi   方法   example   init   and   分析   root   实现   nod   

原文地址:http://www.cnblogs.com/beiyeqingteng/p/6064106.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!