144. 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?

##Approach 1: recursive

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        preorderHelper(root, res);
        return res;
    }
    
    private void preorderHelper(TreeNode node, List<Integer> res) {
        if(node == null) return;
        res.add(node.val);
        preorderHelper(node.left, res);
        preorderHelper(node.right, res);
    }
}

##Approach 2: iterative

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root == null) return res;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        while(!stack.empty()) {
            TreeNode node = stack.pop();
            res.add(node.val);
            if(node.right != null) stack.push(node.right);
            if(node.left != null) stack.push(node.left);
        }
        return res;
    }
}