Two Sum in BST in DSA Typescript - Build from Scratch

Two Sum in BST

📖 Scenario: You are working with a Binary Search Tree (BST) that stores numbers. You want to find out if there are two numbers in the BST that add up to a specific target value. This is useful in many real-world cases like checking if two prices in a sorted product list sum up to a gift card amount.

🎯 Goal: Build a program that checks if there exist two nodes in the BST whose values add up to a given target number.

📋 What You'll Learn

Create a BST with the exact nodes given.

Create a target number variable.

Write a function to find if two numbers in the BST sum to the target.

Print true if such a pair exists, otherwise false.

💡 Why This Matters

🌍 Real World

Finding pairs of numbers that sum to a target is common in finance, shopping discounts, and data analysis where data is stored in sorted structures like BSTs.

💼 Career

Understanding tree traversal and set usage is important for software engineers working with search trees, databases, and optimization problems.

1

Create the BST nodes

Create a class called TreeNode with val, left, and right properties. Then create the BST with these exact nodes and structure:
root node with value 5,
left child with value 3,
right child with value 6,
left child's left child with value 2,
left child's right child with value 4,
right child's right child with value 7.

DSA Typescript

// Define TreeNode class and create the BST nodes as described
// Your code here

Hint

Start by defining the TreeNode class with a constructor. Then create the root node and link its children exactly as described.

2

Set the target sum value

Create a variable called target and set it to the number 9.

DSA Typescript

class TreeNode {
    val: number;
    left: TreeNode | null;
    right: TreeNode | null;
    constructor(val: number, left: TreeNode | null = null, right: TreeNode | null = null) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

const root = new TreeNode(5,
    new TreeNode(3,
        new TreeNode(2),
        new TreeNode(4)
    ),
    new TreeNode(6,
        null,
        new TreeNode(7)
    )
);

// Create a variable called target and set it to 9
// Your code here

Hint

Just create a variable named target and assign it the value 9.

3

Write the function to find two sum in BST

Write a function called findTarget that takes root of type TreeNode | null and k of type number. Use a helper set called seen to store visited node values. Traverse the BST using a helper function dfs. For each node, check if k - node.val is in seen. If yes, return true. Otherwise, add node.val to seen and continue. Return false if no pair found.

DSA Typescript

class TreeNode {
    val: number;
    left: TreeNode | null;
    right: TreeNode | null;
    constructor(val: number, left: TreeNode | null = null, right: TreeNode | null = null) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

const root = new TreeNode(5,
    new TreeNode(3,
        new TreeNode(2),
        new TreeNode(4)
    ),
    new TreeNode(6,
        null,
        new TreeNode(7)
    )
);

const target = 9;

// Write the function findTarget(root: TreeNode | null, k: number): boolean
// Your code here

Hint

Use a set to remember values you have seen. Traverse the tree with a helper function. For each node, check if the complement (target - node value) is in the set. If yes, return true. Otherwise, add the node value to the set and continue.

4

Print the result of two sum check

Write a console.log statement to print the result of calling findTarget with root and target.

DSA Typescript

class TreeNode {
    val: number;
    left: TreeNode | null;
    right: TreeNode | null;
    constructor(val: number, left: TreeNode | null = null, right: TreeNode | null = null) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

const root = new TreeNode(5,
    new TreeNode(3,
        new TreeNode(2),
        new TreeNode(4)
    ),
    new TreeNode(6,
        null,
        new TreeNode(7)
    )
);

const target = 9;

function findTarget(root: TreeNode | null, k: number): boolean {
    const seen = new Set();
    function dfs(node: TreeNode | null): boolean {
        if (!node) return false;
        if (seen.has(k - node.val)) {
            return true;
        }
        seen.add(node.val);
        return dfs(node.left) || dfs(node.right);
    }
    return dfs(root);
}

// Print the result of findTarget(root, target)
// Your code here

Hint

Call findTarget with root and target, then print the result using console.log.

Practice

1. What is the main goal of the Two Sum in BST problem?

easy

A. Check if a BST is balanced

B. Find two nodes in a BST whose values add up to a target sum

C. Count the number of nodes in a BST

D. Find the maximum value in a BST

Solution

1. Step 1: Understand the problem statement

   The problem asks to find two nodes in a Binary Search Tree (BST) such that their values add up to a given target sum.

2. Step 2: Identify the goal

   The goal is to check if such a pair exists, not to find maximum, count nodes, or check balance.

3. Final Answer:

   Find two nodes in a BST whose values add up to a target sum -> Option B

4. Quick Check:

   Two Sum in BST = Find pair with target sum [OK]

Hint: Look for two values adding to target in BST [OK]

Common Mistakes:

• Confusing with finding max or min value
• Thinking it counts nodes instead of finding pairs
• Assuming it checks tree balance

2. Which TypeScript syntax correctly defines a BST node for the Two Sum problem?

easy

A. interface TreeNode { val: string; left: TreeNode; right: TreeNode; }

B. function TreeNode(val) { this.val = val; this.left = null; this.right = null; }

C. type TreeNode = { val: number; left?: TreeNode; right?: TreeNode; }

D. class TreeNode { val: number; left: TreeNode | null; right: TreeNode | null; constructor(val: number) { this.val = val; this.left = null; this.right = null; } }

Solution

1. Step 1: Check correct TypeScript class syntax for BST node

   class TreeNode { val: number; left: TreeNode | null; right: TreeNode | null; constructor(val: number) { this.val = val; this.left = null; this.right = null; } } defines a class with number value and nullable left/right children, matching BST node structure.

2. Step 2: Compare other options

   function TreeNode(val) { this.val = val; this.left = null; this.right = null; } is JavaScript function style, not TypeScript class. interface TreeNode { val: string; left: TreeNode; right: TreeNode; } uses string type for val and non-nullable children, which is incorrect. type TreeNode = { val: number; left?: TreeNode; right?: TreeNode; } uses optional properties but lacks constructor and proper null handling.

3. Final Answer:

   class TreeNode { val: number; left: TreeNode | null; right: TreeNode | null; constructor(val: number) { this.val = val; this.left = null; this.right = null; } } -> Option D

4. Quick Check:

   Correct BST node class syntax = class TreeNode { val: number; left: TreeNode | null; right: TreeNode | null; constructor(val: number) { this.val = val; this.left = null; this.right = null; } } [OK]

Hint: Use class with nullable left/right and number val [OK]

Common Mistakes:

• Using string type for node value
• Not initializing left and right to null
• Using JavaScript function instead of TypeScript class

3. Given the BST below and target = 9, what will the Two Sum function return?

    5
   / \
  3   7
 /     \
2       8

Function uses in-order traversal and a set to find pairs.

medium

A. true

B. false

C. null

D. Error

Solution

1. Step 1: Traverse BST in-order and check pairs

   In-order traversal gives sorted values: [2, 3, 5, 7, 8]. We check if target - current value exists in set.

2. Step 2: Find pair summing to 9

   Check pairs: 2 + 7 = 9 (found), so function returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   2 + 7 = 9 found in BST [OK]

Hint: In-order traversal gives sorted list to find pairs [OK]

Common Mistakes:

• Assuming no pair without checking all nodes
• Confusing traversal order
• Ignoring set lookup for complements

4. Identify the bug in this TypeScript Two Sum in BST function snippet:

function findTarget(root: TreeNode | null, k: number): boolean {
  const set = new Set<number>();
  function dfs(node: TreeNode | null): boolean {
    if (!node) return false;
    if (set.has(k - node.val)) return true;
    set.add(node.val);
    dfs(node.left);
    dfs(node.right);
    return false;
  }
  return dfs(root);
}

medium

A. The base case should return true when node is null

B. Set is not initialized properly

C. The function returns false too early, missing true from recursive calls

D. The function should use a queue instead of recursion

Solution

1. Step 1: Analyze recursive calls and return values

   The dfs function calls dfs(node.left) and dfs(node.right) but ignores their return values, always returning false at the end.

2. Step 2: Identify missing return propagation

   If dfs(node.left) or dfs(node.right) returns true, dfs should return true immediately to propagate the success.

3. Final Answer:

   The function returns false too early, missing true from recursive calls -> Option C

4. Quick Check:

   Return recursive true results to avoid missing pairs [OK]

Hint: Return recursive calls' results to propagate true [OK]

Common Mistakes:

• Ignoring return values of recursive calls
• Returning false before checking all nodes
• Misunderstanding base case return

5. You want to optimize Two Sum in BST for large trees. Which approach best reduces time complexity while using O(h) space, where h is tree height?

hard

A. Use two pointers with iterative in-order and reverse in-order traversals simultaneously

B. Convert BST to array and use nested loops to find pairs

C. Use breadth-first search with a hash set to find pairs

D. Perform recursive DFS and store all node values in a list before checking pairs

Solution

1. Step 1: Understand space and time constraints

   We want to reduce time complexity and keep space to O(h), where h is tree height.

2. Step 2: Evaluate approaches

   Use two pointers with iterative in-order and reverse in-order traversals simultaneously uses two pointers with iterative in-order (smallest to largest) and reverse in-order (largest to smallest) traversals, allowing pair search in O(n) time and O(h) space. Other options use extra space O(n) or nested loops with O(n^2) time.

3. Final Answer:

   Use two pointers with iterative in-order and reverse in-order traversals simultaneously -> Option A

4. Quick Check:

   Two pointer BST traversal = O(n) time, O(h) space [OK]

Hint: Use two pointers with in-order and reverse in-order iterators [OK]

Common Mistakes:

• Using nested loops causing O(n^2) time
• Storing all nodes in list increasing space to O(n)
• Using BFS which uses more space than O(h)