leetcode------Binary Tree Postorder Traversal

标题: Binary Tree Postorder Traversal
通过率: 31.8%
难度:

 

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?

二叉树的后续遍历,代码如下:

 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        ArrayList<Integer> result=new ArrayList<Integer>();
13         vistTree(result,root);
14         return result;
15     }
16     public void vistTree(List<Integer> temp,TreeNode root){
17         if(root!=null){
18             vistTree(temp,root.left);
19             
20             vistTree(temp,root.right);
21             temp.add(root.val);
22         }
23     }
24 }

 

posted @ 2015-03-03 16:54  pku_smile  阅读(95)  评论(0编辑  收藏  举报