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[Leetcode] Binary Tree Postorder Traversal

时间:2014-10-22 14:19:54      阅读:123      评论:0      收藏:0      [点我收藏+]

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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

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

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