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?

Using Stack

/**
 * 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> result = new ArrayList<Integer>();
        if(root == null)
            return result;
        Stack<TreeNode> s = new Stack<TreeNode>();
        s.add(root);
        while(!s.isEmpty()){
            TreeNode tmp = s.pop();
            result.add(tmp.val);
            if(tmp.right != null){
                s.push(tmp.right);
            }
            if(tmp.left != null){
                s.push(tmp.left);
            }
        }
        
        return result;
    }
}

Using Recursion

/**
 * 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> result = new ArrayList<Integer>();
        if(root == null)
            return result;
        result.add(root.val);
        if(root.left != null)
            result.addAll(preorderTraversal(root.left));
        if(root.right != null)
            result.addAll(preorderTraversal(root.right));
        return result;
    }
}

DP

public ArrayList<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        
        preorder(root, result);
        
        return result;
    }
    
    public void preorder(TreeNode root, ArrayList<Integer> result){
        if(root != null){
            result.add(root.val);
            preorder(root.left, result);
            preorder(root.right, result);
        }
    }