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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1301_1400.s1382_balance_a_binary_search_tree;

// #Medium #Depth_First_Search #Greedy #Tree #Binary_Tree #Binary_Search_Tree #Divide_and_Conquer
// #2022_03_21_Time_5_ms_(65.24%)_Space_63_MB_(9.20%)

import com_github_leetcode.TreeNode;
import java.util.ArrayList;
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;
 *     }
 * }
 */
/**
 * 1382 - Balance a Binary Search Tree\.
 *
 * Medium
 *
 * Given the `root` of a binary search tree, return _a **balanced** binary search tree with the same node values_. If there is more than one answer, return **any of them**.
 *
 * A binary search tree is **balanced** if the depth of the two subtrees of every node never differs by more than `1`.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/08/10/balance1-tree.jpg)
 *
 * **Input:** root = [1,null,2,null,3,null,4,null,null]
 *
 * **Output:** [2,1,3,null,null,null,4]
 *
 * **Explanation:** This is not the only correct answer, [3,1,4,null,2] is also correct. 
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/08/10/balanced2-tree.jpg)
 *
 * **Input:** root = [2,1,3]
 *
 * **Output:** [2,1,3] 
 *
 * **Constraints:**
 *
 * *   The number of nodes in the tree is in the range [1, 10⁴].
 * *   1 <= Node.val <= 10⁵
**/
public class Solution {
    public TreeNode balanceBST(TreeNode root) {
        List inorder = inorder(root, new ArrayList<>());
        return dfs(inorder, 0, inorder.size() - 1);
    }

    private List inorder(TreeNode root, List list) {
        if (root == null) {
            return list;
        }
        inorder(root.left, list);
        list.add(root.val);
        return inorder(root.right, list);
    }

    private TreeNode dfs(List nums, int start, int end) {
        if (end < start) {
            return null;
        }
        int mid = (start + end) / 2;
        TreeNode root = new TreeNode(nums.get(mid));
        root.left = dfs(nums, start, mid - 1);
        root.right = dfs(nums, mid + 1, end);
        return root;
    }
}