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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0801_0900.s0897_increasing_order_search_tree;

// #Easy #Depth_First_Search #Tree #Binary_Tree #Stack #Binary_Search_Tree
// #2022_03_28_Time_0_ms_(100.00%)_Space_39.9_MB_(83.16%)

import com_github_leetcode.TreeNode;
import java.util.LinkedList;
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;
 *     }
 * }
 */
/**
 * 897 - Increasing Order Search Tree\.
 *
 * Easy
 *
 * Given the `root` of a binary search tree, rearrange the tree in **in-order** so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/11/17/ex1.jpg)
 *
 * **Input:** root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
 *
 * **Output:** [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9] 
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/11/17/ex2.jpg)
 *
 * **Input:** root = [5,1,7]
 *
 * **Output:** [1,null,5,null,7] 
 *
 * **Constraints:**
 *
 * *   The number of nodes in the given tree will be in the range `[1, 100]`.
 * *   `0 <= Node.val <= 1000`
**/
public class Solution {
    public TreeNode increasingBST(final TreeNode root) {
        final List list = new LinkedList<>();
        traverse(root, list);
        for (int i = 1; i < list.size(); i++) {
            list.get(i - 1).right = list.get(i);
            list.get(i).left = null;
        }
        return list.get(0);
    }

    private void traverse(final TreeNode root, final List list) {
        if (root != null) {
            traverse(root.left, list);
            list.add(root);
            traverse(root.right, list);
        }
    }
}