LeetCode #938 — EASY

Range Sum of BST

Build confidence with an intuition-first walkthrough focused on tree fundamentals.

The Problem

Problem Statement

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:

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.

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.

Step 01

Brute Force Baseline

Problem summary: 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].

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

[10,5,15,3,7,null,18]
7
15

Example 2

[10,5,15,3,7,13,18,1,null,6]
6
10

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #938: Range Sum of BST
/**
 * 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) {
        return dfs(root, low, high);
    }

    private int dfs(TreeNode root, int low, int high) {
        if (root == null) {
            return 0;
        }
        int x = root.val;
        int ans = low <= x && x <= high ? x : 0;
        if (x > low) {
            ans += dfs(root.left, low, high);
        }
        if (x < high) {
            ans += dfs(root.right, low, high);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #938: Range Sum of BST
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func rangeSumBST(root *TreeNode, low int, high int) int {
	var dfs func(*TreeNode) int
	dfs = func(root *TreeNode) (ans int) {
		if root == nil {
			return 0
		}
		x := root.Val
		if low <= x && x <= high {
			ans += x
		}
		if x > low {
			ans += dfs(root.Left)
		}
		if x < high {
			ans += dfs(root.Right)
		}
		return
	}
	return dfs(root)
}

# Accepted solution for LeetCode #938: Range Sum of BST
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rangeSumBST(self, root: Optional[TreeNode], low: int, high: int) -> int:
        def dfs(root: Optional[TreeNode]) -> int:
            if root is None:
                return 0
            x = root.val
            ans = x if low <= x <= high else 0
            if x > low:
                ans += dfs(root.left)
            if x < high:
                ans += dfs(root.right)
            return ans

        return dfs(root)

// Accepted solution for LeetCode #938: Range Sum of BST
struct Solution;
use rustgym_util::*;

trait Preorder {
    fn preorder(&self, l: i32, r: i32, sum: &mut i32);
}

impl Preorder for TreeLink {
    fn preorder(&self, l: i32, r: i32, sum: &mut i32) {
        if let Some(node) = self {
            let node = node.borrow();
            let left = &node.left;
            let right = &node.right;
            let val = node.val;
            if val >= l && val <= r {
                *sum += val;
            }
            if val > l {
                left.preorder(l, r, sum)
            }
            if val < r {
                right.preorder(l, r, sum);
            }
        }
    }
}

impl Solution {
    fn range_sum_bst(root: TreeLink, l: i32, r: i32) -> i32 {
        let mut sum = 0;
        root.preorder(l, r, &mut sum);
        sum
    }
}

#[test]
fn test() {
    let root = tree!(10, tree!(5, tree!(3), tree!(7)), tree!(15, None, tree!(18)));
    assert_eq!(Solution::range_sum_bst(root, 7, 15), 32);
}

// Accepted solution for LeetCode #938: Range Sum of BST
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function rangeSumBST(root: TreeNode | null, low: number, high: number): number {
    const dfs = (root: TreeNode | null): number => {
        if (!root) {
            return 0;
        }
        const { val, left, right } = root;
        let ans = low <= val && val <= high ? val : 0;
        if (val > low) {
            ans += dfs(left);
        }
        if (val < high) {
            ans += dfs(right);
        }
        return ans;
    };
    return dfs(root);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.