Range Sum of BST easy

Problem Statement

Given the root node of a binary search tree (BST) 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:

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:

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

Steps and Explanation

The problem can be efficiently solved using a recursive approach that leverages the properties of a BST. Since a BST is ordered, we can prune our search significantly.

1. Recursive Function: We define a recursive function rangeSumBST that takes the root node, low, and high as input.

2. Base Case: If the current node is null, we return 0 (no sum).

3. Pruning:

   • If the current node's value is less than low, we only need to recursively search the right subtree (since values in the left subtree will be even smaller).
   • If the current node's value is greater than high, we only need to recursively search the left subtree (since values in the right subtree will be even larger).

4. Inclusion: If the current node's value is within the range [low, high], we include its value in the sum and recursively search both the left and right subtrees.

5. Return Value: The function returns the sum of the current node's value (if included) and the recursive sums from the left and right subtrees.

Code

/**
 * 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;
 *     }
 * }
 */
class Solution {
    public int rangeSumBST(TreeNode root, int low, int high) {
        if (root == null) {
            return 0;
        }

        if (root.val < low) {
            return rangeSumBST(root.right, low, high);
        } else if (root.val > high) {
            return rangeSumBST(root.left, low, high);
        } else {
            return root.val + rangeSumBST(root.left, low, high) + rangeSumBST(root.right, low, high);
        }
    }
}

Complexity

• Time Complexity: O(N) in the worst case, where N is the number of nodes in the BST. However, due to pruning, it's often significantly less than O(N) if the range [low, high] is small or strategically located within the tree.
• Space Complexity: O(H) in the worst case, where H is the height of the BST. This is due to the recursive call stack. In a balanced BST, H is log(N), but in a skewed BST, H can be N.