首页 > 其他 > 详细

[Leetcode] Binary Tree Postorder Traversal

时间:2014-10-22 14:18:19      阅读:183      评论:0      收藏:0      [点我收藏+]

Given a binary tree, return the postorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

 

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

 

Solution 1: 非递归

 

 

Solution 2: 递归

 1 /**
 2  * Definition for binary tree
 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 List<Integer> postorderTraversal(TreeNode root) {
12         List<Integer> result=new ArrayList<Integer>();
13         myPostorderTraversal(root,result);
14         return result;
15     }
16 
17     private void myPostorderTraversal(TreeNode root, List<Integer> result) {
18         // TODO Auto-generated method stub
19         if(root!=null){
20             myPostorderTraversal(root.left, result);
21             myPostorderTraversal(root.right, result);
22             result.add(root.val);
23         }
24     }
25 }

 

[Leetcode] Binary Tree Postorder Traversal

原文:http://www.cnblogs.com/Phoebe815/p/4042933.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!