Range Sum of BST

LeetCode 975 | Difficulty: Easy

Easy

Problem Description

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 * 10^4]`.

- `1 <= Node.val <= 10^5`

- `1 <= low <= high <= 10^5`

- All `Node.val` are **unique**.

Topics: Tree, Depth-First Search, Binary Search Tree, Binary Tree

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Approach

Tree DFS

Traverse the tree recursively (or with a stack). At each node, decide: what information do I need from the left/right subtrees? Process: go left → go right → combine results. Consider preorder, inorder, or postorder traversal based on when you need to process the node.

When to use

Path problems, subtree properties, tree structure manipulation.

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Solutions

Solution 1: C# (Best: 184 ms)

Metric	Value
Runtime	184 ms
Memory	46.7 MB
Date	2022-01-03

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int result = 0;
    
    public int RangeSumBST(TreeNode root, int low, int high) {
    if(root == null) return 0;
    int val = root.val;
    if(val>=low  && val <= high) result += val;;
    
    RangeSumBST(root.left, low, high);
    RangeSumBST(root.right, low, high);
    return result;
}
}

📜 3 more C# submission(s)

Submission (2022-01-03) — 184 ms, 46.8 MB

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int RangeSumBST(TreeNode root, int low, int high) {
        int[] result = new int[1];
    RangeSumBSTHelper(root, low, high, result);
    return result[0];
}

public void RangeSumBSTHelper(TreeNode root, int low, int high, int[] result)
{
    if(root == null) return;
    int val = root.val;
    if(val>=low  && val <= high) result[0] += val;;
    
    RangeSumBSTHelper(root.left, low, high, result);
    RangeSumBSTHelper(root.right, low, high, result);
}
}

Submission (2022-01-03) — 192 ms, 46.7 MB

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int RangeSumBST(TreeNode root, int low, int high) {
        int[] result = new int[1];
    RangeSumBSTHelper(root, low, high, result);
    return result[0];
}

public void RangeSumBSTHelper(TreeNode root, int low, int high, int[] result)
{
    if(root == null) return;
    int val = root.val;
    if(val>=low  && val <= high) result[0] += val;;
    if(val >= low)
    RangeSumBSTHelper(root.left, low, high, result);
    if(val <= high)
    RangeSumBSTHelper(root.right, low, high, result);
}
}

Submission (2022-01-03) — 356 ms, 46.6 MB

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int result = 0;
    
    public int RangeSumBST(TreeNode root, int low, int high) {
    if(root == null) return 0;
    int val = root.val;
    if(val>=low  && val <= high) result += val;;
    if(val>=low)
    RangeSumBST(root.left, low, high);
        if(val <= high)
    RangeSumBST(root.right, low, high);
    return result;
}
}

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Complexity Analysis

Approach	Time	Space
Tree Traversal	$O(n)$	$O(h)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• Consider: "What information do I need from each subtree?" — this defines your recursive return value.