Binary Tree Postorder Traversal

Link: http://oj.leetcode.com/problems/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?

Iterative version:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        ArrayList<Integer> nodes = new ArrayList<Integer>();
        if (root == null)
            return nodes;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode node;
        stack.push(root);
        while (!stack.isEmpty()) {
            node = stack.pop();
            nodes.add(node.val);
            if (node.left != null)
                stack.push(node.left);
            if (node.right != null)
                stack.push(node.right);
        }
        ArrayList<Integer> result = new ArrayList<Integer>();
        for (int i = nodes.size() - 1; i >= 0; i--) {
            result.add(nodes.get(i));
        }
        return result;
    }
}

Recursion version:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
   ArrayList<Integer> nodes = new ArrayList<Integer>();

    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        postorderTraversal_recursion(root);
        return nodes;
    }

    public void postorderTraversal_recursion(TreeNode root) {
        if (root == null)
            return;
        postorderTraversal_recursion(root.left);
        postorderTraversal_recursion(root.right);
        nodes.add(root.val);
    }
}