Two Sum in BST in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Two Sum in BST with target 9

What is the output of the following code that checks if there exist two nodes in the BST whose values sum to 9?

DSA Javascript

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

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

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

console.log(findTarget(root, 9));

A TypeError

B false

C undefined

D true

💡 Hint

Think about whether any two nodes add up to 9 in the BST.

The nodes with values 2 and 7 sum to 9, so the function returns true.

❓ Predict Output

intermediate

Output of Two Sum in BST with target 28

What is the output of the following code that checks if there exist two nodes in the BST whose values sum to 28?

DSA Javascript

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

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

const root = new TreeNode(10,
  new TreeNode(5,
    new TreeNode(3),
    new TreeNode(7)
  ),
  new TreeNode(15,
    null,
    new TreeNode(18)
  )
);

console.log(findTarget(root, 28));

A true

B false

C null

D ReferenceError

💡 Hint

Check if any two nodes add up to 28 in the BST.

The nodes with values 10 and 18 sum to 28, so the function returns true.

🔧 Debug

advanced

Identify the error in Two Sum in BST implementation

What error will this code produce when run?

DSA Javascript

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

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

const root = new TreeNode(1,
  new TreeNode(0),
  new TreeNode(3)
);

console.log(findTarget(root, "3"));

A No error, outputs true

B true

C false

D TypeError

💡 Hint

Consider JavaScript type coercion in arithmetic operations.

No error is produced. JavaScript coerces the string "3" to number 3 during subtraction, and finds nodes 0 and 3 sum to 3, outputting true.

🧠 Conceptual

advanced

Why use a Set in Two Sum in BST?

Why does the Two Sum in BST algorithm use a Set to store node values during traversal?

A To count the frequency of each node value

B To sort the node values for binary search

C To keep track of visited nodes and check complements efficiently

D To store nodes in insertion order for traversal

💡 Hint

Think about how to quickly check if a complement value exists.

The Set allows O(1) lookup to check if the complement (target - current node value) exists among visited nodes.

🚀 Application

expert

Number of pairs summing to target in BST

Given the BST below, how many unique pairs of nodes sum to 10? BST structure: 5 / \ 3 7 / \ \ 2 4 8

A 1

B 2

C 3

D 4

💡 Hint

List all pairs and count those summing to 10.

Pairs (2,8) and (3,7) sum to 10. The pair (4,6) is invalid as 6 is not in the tree. Since 5 appears once, (5,5) is not valid. So total 2 pairs.

Practice

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

easy

A. Find the maximum value in the BST.

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

C. Count the total number of nodes in the BST.

D. Check if the BST is balanced.

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 max value, count nodes, or check balance.

3. Final Answer:

   Find two nodes in the 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 pairs adding to target in BST nodes [OK]

Common Mistakes:

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

2. Which of the following is the correct way to declare a BST node in JavaScript for the Two Sum problem?

easy

A. function Node(val) { this.val = val; this.left = null; this.right = null; }

B. let Node = (val) => { val = val; left = null; right = null; }

C. class Node { constructor(val) { val = val; left = null; right = null; } }

D. var Node = function(val) { this.value = val; this.left = this.right = undefined; }

Solution

1. Step 1: Check correct property assignments

   function Node(val) { this.val = val; this.left = null; this.right = null; } correctly assigns this.val, this.left, and this.right to initialize a BST node.

2. Step 2: Identify syntax and property errors in other options

   let Node = (val) => { val = val; left = null; right = null; } misses this keyword and uses arrow function incorrectly. class Node { constructor(val) { val = val; left = null; right = null; } } assigns to local variables, not this. var Node = function(val) { this.value = val; this.left = this.right = undefined; } uses this.value instead of this.val and sets children to undefined instead of null.

3. Final Answer:

   function Node(val) { this.val = val; this.left = null; this.right = null; } -> Option A

4. Quick Check:

   Correct node declaration uses this.val and null children [OK]

Hint: Use function with this.val, this.left, this.right = null [OK]

Common Mistakes:

• Forgetting 'this' keyword for properties
• Using arrow functions for constructor
• Assigning children to undefined instead of null

3. Given the BST below and target = 9, what will the function findTarget(root, 9) return?

    5
   / \
  3   6
 / \   \
2   4   7

Assuming findTarget returns true if two nodes sum to target, else false.

medium

A. true

B. false

C. null

D. undefined

Solution

1. Step 1: Identify pairs that sum to 9

   Check pairs: 2+7=9, 3+6=9, 4+5=9. All these pairs exist in the BST.

2. Step 2: Confirm function returns true if any pair matches

   Since pairs exist, findTarget returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Pairs sum to 9 = true [OK]

Hint: Check pairs adding to target in BST nodes [OK]

Common Mistakes:

• Ignoring valid pairs in BST
• Assuming function returns node instead of boolean
• Confusing null or undefined as return

4. Consider this buggy code snippet for Two Sum in BST:

function findTarget(root, k) {
  const set = new Set();
  function dfs(node) {
    if (!node) return false;
    if (set.has(k - node.val)) return true;
    set.add(node.val);
    dfs(node.left);
    dfs(node.right);
  }
  return dfs(root);
}

What is the main bug causing incorrect results?

medium

A. Using Set instead of Array for storage

B. Checking for k - node.val before adding node.val

C. Missing return statements in recursive calls inside dfs

D. Not handling empty tree case

Solution

1. Step 1: Analyze recursive calls in dfs

   The recursive calls to dfs(node.left) and dfs(node.right) do not return their results, so true found deeper is ignored.

2. Step 2: Fix by returning recursive calls

   Adding return before recursive calls ensures if true is found, it propagates up and final result is correct.

3. Final Answer:

   Missing return statements in recursive calls inside dfs -> Option C

4. Quick Check:

   Return recursive calls to propagate true [OK]

Hint: Always return recursive calls to propagate results [OK]

Common Mistakes:

• Not returning recursive call results
• Confusing Set usage as bug
• Ignoring base case handling

5. You want to optimize the Two Sum in BST problem to use O(h) space, where h is tree height. Which approach is best?

hard

A. Convert BST to a linked list and then find pairs.

B. Store all node values in an array and use a hash set to find pairs.

C. Perform a breadth-first search and check pairs at each level.

D. Use two iterators: one in-order and one reverse in-order traversal simultaneously.

Solution

1. Step 1: Understand space complexity of approaches

   Storing all nodes uses O(n) space. BFS also uses O(n) space. Converting to linked list uses O(n).

2. Step 2: Use two iterators approach

   Using two iterators (one in-order for smallest, one reverse in-order for largest) uses O(h) space, where h is tree height, optimizing space.

3. Final Answer:

   Use two iterators: one in-order and one reverse in-order traversal simultaneously. -> Option D

4. Quick Check:

   Two iterators = O(h) space optimization [OK]

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

Common Mistakes:

• Using full array storage increasing space to O(n)
• Trying BFS which uses more space
• Converting BST to linked list wastes space