LeetCode Medium: Binary Tree Preorder Traversal

Hello Fellas,

Here are 2 of my accepted solution approaches (both O(n) where n is the number of nodes in the input tree) for solving the Problem #144 Binary Tree Preorder Traversal on Leetcode:

Approach 1: Simple Recursive

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

Approach 2: Iterative Stack based

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

Know of more ways, feel free to share your code or suggestions in comments below!