Binary Tree Preorder Traversal

Link: http://oj.leetcode.com/problems/binary-tree-preorder-traversal/

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    
     2
    /
   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

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> preorderTraversal(TreeNode root) {

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

    }
}

Iteration 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> preorderTraversal(TreeNode root) {
    ArrayList<Integer> nodes = new ArrayList<Integer>();
        if (root == null)
            return nodes;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        // this is used to identify the node pop from stack
        TreeNode node;
        stack.push(root);
        while (!stack.isEmpty()) {
            node = stack.pop();
            nodes.add(node.val);
            if (node.right != null)
                stack.push(node.right);
            if (node.left != null)
                stack.push(node.left);
        }
        return nodes;
    }
}