Two Sum in BST in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Two Sum problem in a Binary Search Tree (BST)?

To find if there exist two nodes in the BST such that their values add up to a given target sum.

intermediate

Why is BST property useful in solving the Two Sum problem efficiently?

Because BST nodes are ordered, we can use two pointers (one from smallest, one from largest) to find the target sum without checking all pairs.

beginner

What data structure can be used to store visited node values to find the complement for Two Sum in BST?

A hash set can be used to store visited values and quickly check if the complement (target - current node value) exists.

advanced

Explain the two-pointer approach for Two Sum in BST.

Use an iterator to get the smallest value (inorder traversal) and another iterator for the largest value (reverse inorder traversal). Move pointers inward based on sum comparison with target.

intermediate

What is the time complexity of the Two Sum in BST using the two-pointer approach?

O(n), where n is the number of nodes, because each node is visited at most once by the two iterators.

What traversal is used to get nodes in ascending order in a BST?

A Level order traversal

B Preorder traversal

C Postorder traversal

D Inorder traversal

Inorder traversal visits nodes in ascending order for BST.

Which data structure helps check if a complement value exists quickly during Two Sum in BST?

A Queue

B Stack

C Hash set

D Linked list

Hash set allows O(1) average time lookup for complements.

In the two-pointer approach for Two Sum in BST, what does moving the left pointer forward mean?

A Checking larger values

B Checking random values

C Checking smaller values

D Stopping the search

Moving left pointer forward moves to larger values in ascending order.

What is the space complexity of using a hash set for Two Sum in BST?

A O(1)

B O(n)

C O(log n)

D O(n^2)

Hash set can store up to all node values, so O(n) space.

Which of these is NOT a valid approach to solve Two Sum in BST?

A Using binary search on unsorted array

B Using two pointers from smallest and largest nodes

C Brute force checking all pairs without BST property

D Using hash set to store visited values

Binary search requires sorted data; unsorted array is invalid here.

Describe how you would use two pointers to find two nodes in a BST that sum to a target.

Think about smallest and largest values in BST.

Explain how a hash set can help solve the Two Sum problem in a BST during a tree traversal.

Use extra memory to speed up complement lookup.

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