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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0101_0200.s0144_binary_tree_preorder_traversal;

// #Easy #Depth_First_Search #Tree #Binary_Tree #Stack #Data_Structure_I_Day_10_Tree
// #Udemy_Tree_Stack_Queue #2022_06_24_Time_1_ms_(48.38%)_Space_42_MB_(68.46%)

import com_github_leetcode.TreeNode;
import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

/*
 * 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;
 *     }
 * }
 */
/**
 * 144 - Binary Tree Preorder Traversal\.
 *
 * Easy
 *
 * Given the `root` of a binary tree, return _the preorder traversal of its nodes' values_.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/09/15/inorder_1.jpg)
 *
 * **Input:** root = [1,null,2,3]
 *
 * **Output:** [1,2,3] 
 *
 * **Example 2:**
 *
 * **Input:** root = []
 *
 * **Output:** [] 
 *
 * **Example 3:**
 *
 * **Input:** root = [1]
 *
 * **Output:** [1] 
 *
 * **Example 4:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/09/15/inorder_5.jpg)
 *
 * **Input:** root = [1,2]
 *
 * **Output:** [1,2] 
 *
 * **Example 5:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/09/15/inorder_4.jpg)
 *
 * **Input:** root = [1,null,2]
 *
 * **Output:** [1,2] 
 *
 * **Constraints:**
 *
 * *   The number of nodes in the tree is in the range `[0, 100]`.
 * *   `-100 <= Node.val <= 100`
 *
 * **Follow up:** Recursive solution is trivial, could you do it iteratively?
**/
@SuppressWarnings("java:S1149")
public class Solution {
    public List preorderTraversal(TreeNode root) {
        List result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Stack stack = new Stack<>();
        TreeNode current = root;
        while (current != null || !stack.isEmpty()) {
            while (current != null) {
                result.add(current.val);
                stack.push(current.right);
                current = current.left;
            }
            current = stack.pop();
        }
        return result;
    }
}