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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0094_binary_tree_inorder_traversal;

// #Easy #Top_100_Liked_Questions #Top_Interview_Questions #Depth_First_Search #Tree #Binary_Tree
// #Stack #Data_Structure_I_Day_10_Tree #Udemy_Tree_Stack_Queue #Big_O_Time_O(n)_Space_O(n)
// #2022_06_21_Time_0_ms_(100.00%)_Space_42.7_MB_(9.33%)

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

/*
 * 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;
 *     }
 * }
 */
/**
 * 94 - Binary Tree Inorder Traversal\.
 *
 * Easy
 *
 * Given the `root` of a binary tree, return _the inorder 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:** [1,3,2] 
 *
 * **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:** [2,1] 
 *
 * **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?
**/
public class Solution {
    public List inorderTraversal(TreeNode root) {
        if (root == null) {
            return Collections.emptyList();
        }
        List answer = new ArrayList<>();
        inorderTraversal(root, answer);
        return answer;
    }

    public void inorderTraversal(TreeNode root, List answer) {
        if (root == null) {
            return;
        }
        if (root.left != null) {
            inorderTraversal(root.left, answer);
        }
        answer.add(root.val);
        if (root.right != null) {
            inorderTraversal(root.right, answer);
        }
    }
}