Two Sum in BST in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a value into the BST.

DSA C++

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

TreeNode* insertNode(TreeNode* root, int val) {
    if (root == nullptr) {
        return new TreeNode(val);
    }
    if (val < root->val) {
        root->left = insertNode(root->left, [1]);
    } else {
        root->right = insertNode(root->right, val);
    }
    return root;
}

Drag options to blanks, or click blank then click option'

Aval

Broot->val

Cnullptr

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using root->val instead of val causes wrong recursion.

Passing nullptr instead of val breaks insertion.

When inserting, we recursively call insertNode on the left subtree with the value to insert.

2fill in blank

medium

Complete the code to perform inorder traversal and store values in a vector.

DSA C++

void inorderTraversal(TreeNode* root, std::vector<int>& vals) {
    if (root == nullptr) return;
    inorderTraversal(root->left, vals);
    vals.push_back([1]);
    inorderTraversal(root->right, vals);
}

Drag options to blanks, or click blank then click option'

Aroot->val

Bvals

Croot->left

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Adding root instead of root->val causes type errors.

Adding root->left instead of root->val adds wrong data.

In inorder traversal, we add the current node's value to the vector.

3fill in blank

hard

Fix the error in the two-pointer search for the target sum in the sorted vector.

DSA C++

bool findTargetSum(const std::vector<int>& vals, int target) {
    int left = 0, right = vals.size() - 1;
    while (left < right) {
        int sum = vals[left] + vals[[1]];
        if (sum == target) return true;
        else if (sum < target) left++;
        else right--;
    }
    return false;
}

Drag options to blanks, or click blank then click option'

Aleft

Btarget

C0

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using vals[left] twice causes incorrect sums.

Using vals[0] or vals[target] causes out-of-bounds errors.

We add vals[left] and vals[right] to check the sum.

4fill in blank

hard

Fill both blanks to complete the function that checks if the BST has two elements summing to target.

DSA C++

bool findTarget(TreeNode* root, int k) {
    std::vector<int> vals;
    inorderTraversal(root, vals);
    int left = 0, right = vals.size() - 1;
    while (left < right) {
        int sum = vals[left] + vals[right];
        if (sum == k) return true;
        else if (sum [1] k) left++;
        else right[2];
    }
    return false;
}

Drag options to blanks, or click blank then click option'

A<

B>

C--

D++

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < in the first blank reverses logic.

Using ++ on right pointer causes infinite loop.

If sum is less than k, move left pointer up; else move right pointer down.

5fill in blank

hard

Fill all three blanks to build a complete two-sum in BST function using inorder traversal and two-pointer technique.

DSA C++

bool findTarget(TreeNode* root, int k) {
    std::vector<int> vals;
    [1](root, vals);
    int left = 0, right = vals.size() - 1;
    while (left < right) {
        int sum = vals[left] + vals[right];
        if (sum == k) return true;
        else if (sum < k) left[2];
        else right[3];
    }
    return false;
}

Drag options to blanks, or click blank then click option'

AinorderTraversal

B++

C--

DinsertNode

Attempts:

3 left

💡 Hint

Common Mistakes

Calling insertNode instead of inorderTraversal.

Using wrong pointer increments causing infinite loops.

We call inorderTraversal to fill vals, then move pointers correctly to find the sum.

Practice

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

easy

A. Find the maximum value in a BST

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. Check if a BST is balanced

Solution

1. Step 1: Understand the problem statement

   The problem asks to find if there exist 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 for the existence of such a pair, 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 means finding two nodes summing to target [OK]

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

Common Mistakes:

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

2. Which of the following C++ code snippets correctly declares a BST node structure for the Two Sum problem?

easy

A. struct TreeNode { int val; TreeNode left; TreeNode* right; };

B. struct TreeNode { int val; TreeNode left; TreeNode right; };

C. struct TreeNode { int val; TreeNode* left; TreeNode* right; };

D. struct TreeNode { int val; TreeNode* left; TreeNode right; };

Solution

1. Step 1: Check pointer usage in struct

   In C++, child nodes in a tree are pointers to TreeNode objects, so left and right must be pointers.

2. Step 2: Validate each option

   struct TreeNode { int val; TreeNode* left; TreeNode* right; }; uses TreeNode* for left and right, which is correct. Options A, B, and C incorrectly use TreeNode without pointers for one or both children.

3. Final Answer:

   struct TreeNode { int val; TreeNode* left; TreeNode* right; }; -> Option C

4. Quick Check:

   BST nodes use pointers for children [OK]

Hint: Children in BST nodes are pointers in C++ [OK]

Common Mistakes:

• Using objects instead of pointers for children
• Mixing pointer and non-pointer types
• Forgetting semicolon after struct

3. What is the output of the following code snippet that checks for two sum in BST?

struct TreeNode {
  int val;
  TreeNode* left;
  TreeNode* right;
  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

bool findTarget(TreeNode* root, int k) {
  std::unordered_set<int> seen;
  std::function dfs = [&](TreeNode* node) {
    if (!node) return false;
    if (seen.count(k - node->val)) return true;
    seen.insert(node->val);
    return dfs(node->left) || dfs(node->right);
  };
  return dfs(root);
}

int main() {
  TreeNode* root = new TreeNode(5);
  root->left = new TreeNode(3);
  root->right = new TreeNode(6);
  root->left->left = new TreeNode(2);
  root->left->right = new TreeNode(4);
  root->right->right = new TreeNode(7);
  
  bool result = findTarget(root, 9);
  std::cout << (result ? "true" : "false") << std::endl;
  return 0;
}

medium

A. true

B. false

C. Compilation error

D. Runtime error

Solution

1. Step 1: Understand the tree and target

   The BST has nodes 5,3,6,2,4,7. The target sum is 9.

2. Step 2: Trace the DFS and set usage

   DFS visits nodes and checks if (target - current node value) is in the set. For example, when visiting node 6, it checks if 9-6=3 is in set. Node 3 was visited before, so it returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   3 + 6 = 9 found in BST [OK]

Hint: Check if complement exists in visited nodes set [OK]

Common Mistakes:

• Assuming no pair exists without checking all nodes
• Confusing DFS order with BFS
• Ignoring the use of hash set for complements

4. Identify the bug in the following Two Sum in BST code snippet and select the correct fix:

bool findTarget(TreeNode* root, int k) {
  std::unordered_set<int> seen;
  std::function dfs = [&](TreeNode* node) {
    if (!node) return false;
    if (seen.count(k - node->val)) return true;
    dfs(node->left);
    dfs(node->right);
    seen.insert(node->val);
    return false;
  };
  return dfs(root);
}

medium

A. Insert node->val before recursive calls to children

B. Remove the unordered_set and use a vector instead

C. Change return false to return true at the end

D. Add a null check after inserting node->val

Solution

1. Step 1: Analyze the order of operations

   The code inserts node->val into the set after recursive calls, so complements are not found when visiting children.

2. Step 2: Fix the insertion timing

   Insert node->val before recursive calls so that children can find complements in the set.

3. Final Answer:

   Insert node->val before recursive calls to children -> Option A

4. Quick Check:

   Insert before recursion to find complements [OK]

Hint: Insert current value before recursion to detect complements [OK]

Common Mistakes:

• Inserting after recursion causes missed pairs
• Changing return values incorrectly
• Replacing set with vector reduces efficiency

5. Given a BST with nodes [10, 5, 15, 3, 7, null, 18], how would you modify the Two Sum in BST algorithm to handle the case where the target sum is twice the value of a single node (e.g., target = 14 and node value = 7)?

hard

A. Only check leaf nodes for pairs

B. Ignore pairs where complement equals current node value

C. Use a sorted array instead of a hash set

D. Check if the complement equals the current node value and ensure the node appears twice

Solution

1. Step 1: Understand the problem with duplicates

   If target is twice a node's value, we must ensure two distinct nodes with that value exist.

2. Step 2: Modify algorithm to handle duplicates

   When complement equals current node value, check if the node appears twice (e.g., by counting occurrences or checking different nodes).

3. Final Answer:

   Check if the complement equals the current node value and ensure the node appears twice -> Option D

4. Quick Check:

   Handle duplicates by verifying two distinct nodes [OK]

Hint: Verify two distinct nodes if complement equals current value [OK]

Common Mistakes:

• Ignoring duplicate values causing false positives
• Skipping pairs with same node value
• Switching data structures unnecessarily