[Leetcode] Binary Tree Postorder Traversal

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

 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         if (root == null)
14             return result;
15         Stack<TreeNode> s = new Stack<TreeNode>();
16         TreeNode node = root, prev = root;
17         while (node != null || !s.isEmpty()) {
18             while (node != null) {
19                 s.add(node);
20                 node = node.left;
21             }
22             if (s.size() > 0) {
23                 TreeNode temp = s.peek().right;
24                 if(temp!=null&&temp!=prev){
25                     node=temp;
26                 }else if(temp==prev||temp==null){
27                     node=s.pop();
28                     result.add(node.val);
29                     prev=node;
30                     node=null;
31                 }
32             }
33         }
34         return result;
35     }
36 }

 

 

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 }

 

posted @ 2014-10-22 13:43  Phoebe815  阅读(151)  评论(0)  编辑  收藏  举报