Binary Tree Preorder Tranversal

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?

Solution 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> ans = new ArrayList<>();
        preorderTraversal(root, ans);
        return ans;
    }
    public void preorderTraversal(TreeNode root, List<Integer> list) {
        if(root==null) return;
        list.add(root.val);
        preorderTraversal(root.left, list);
        preorderTraversal(root.right, list);
    }
}

Solution 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> ans = new ArrayList<>();
        Stack<TreeNode> st = new Stack<>();
        while(root!=null || !st.isEmpty()){
            while(root!=null){
                ans.add(root.val);
                st.push(root);
                root = root.left;
            }
            root = st.pop().right;
        }
        return ans;
    }
}