Binary Tree Postorder Traversal

Problem:

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].

Analysis:
The most difficult one among preorder, inorder and postorder traversals. The difficulty lays in that after going down to left it will go right, storing the parent node is a problem. Use one more stack to assist push nodes in reserve order of postorder to the final stack.
Steps:
1.Push the root node to the temp stack.
2.Pop node c from temp stack.
3.If c has left child push it to final stack.If c has right child push it to final stack.
4.Push c to final stack.
5.Repeat 2-4 until temp stack is empty.
6.Print final stack.

Solution:

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 ArrayList<Integer> postorderTraversal(TreeNode root) {  
12:      ArrayList<Integer> result=new ArrayList<Integer>();  
13:      if(root==null)  
14:       return result;  
15:      Stack<TreeNode> t=new Stack<TreeNode>();//temp stack   
16:      Stack<TreeNode> r=new Stack<TreeNode>();//result stack  
17:      t.push(root);  
18:      while(!t.isEmpty())  
19:      {  
20:       TreeNode current=t.pop();  
21:        if(current.left!=null) t.push(current.left);  
22:        if(current.right!=null) t.push(current.right);  
23:        r.push(current);  
24:      }  
25:      while(!r.isEmpty())  
26:      {  
27:       TreeNode current=r.pop();  
28:        result.add(current.val);  
29:      }  
30:      return result;  
31:    }  
32:  }