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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0901_1000.s0938_range_sum_of_bst;

// #Easy #Depth_First_Search #Tree #Binary_Tree #Binary_Search_Tree #Udemy_Tree_Stack_Queue
// #2022_03_30_Time_0_ms_(100.00%)_Space_64.7_MB_(76.99%)

import com_github_leetcode.TreeNode;

/*
 * 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;
 *     }
 * }
 */
/**
 * 938 - Range Sum of BST\.
 *
 * Easy
 *
 * Given the `root` node of a binary search tree and two integers `low` and `high`, return _the sum of values of all nodes with a value in the **inclusive** range_ `[low, high]`.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/11/05/bst1.jpg)
 *
 * **Input:** root = [10,5,15,3,7,null,18], low = 7, high = 15
 *
 * **Output:** 32
 *
 * **Explanation:** Nodes 7, 10, and 15 are in the range [7, 15]. 7 + 10 + 15 = 32. 
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/11/05/bst2.jpg)
 *
 * **Input:** root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10
 *
 * **Output:** 23
 *
 * **Explanation:** Nodes 6, 7, and 10 are in the range [6, 10]. 6 + 7 + 10 = 23. 
 *
 * **Constraints:**
 *
 * *   The number of nodes in the tree is in the range [1, 2 * 10⁴].
 * *   1 <= Node.val <= 10⁵
 * *   1 <= low <= high <= 10⁵
 * *   All `Node.val` are **unique**.
**/
public class Solution {
    public int rangeSumBST(TreeNode root, int low, int high) {
        if (root == null) {
            return 0;
        }
        if (root.val <= high && root.val >= low) {
            int sum = root.val;
            int left = rangeSumBST(root.left, low, high);
            int right = rangeSumBST(root.right, low, high);
            return sum + left + right;
        } else if (root.val < low) {
            return rangeSumBST(root.right, low, high);
        } else {
            return rangeSumBST(root.left, low, high);
        }
    }
}