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 whose values add up to a given target sum.

beginner

Why is a BST useful for solving the Two Sum problem efficiently?

Because BST nodes are sorted in order, allowing efficient searching and traversal to find pairs that sum to the target.

intermediate

What data structure can be used to keep track of visited node values when solving Two Sum in BST?

A HashSet can be used to store values of visited nodes to quickly check if the complement (target - current node value) exists.

intermediate

Describe the two-pointer approach to solve Two Sum in BST.

Perform an in-order traversal to get sorted values, then use two pointers (start and end) to find if any two values sum to the target.

intermediate

What is the time complexity of solving Two Sum in BST using a HashSet during DFS traversal?

O(n), where n is the number of nodes, because each node is visited once and HashSet lookups are O(1) on average.

What traversal is commonly used to get sorted node values from a BST?

A Level-order traversal

B Pre-order traversal

C Post-order traversal

D In-order traversal

In-order traversal visits BST nodes in ascending order.

Which data structure helps check if a complement value exists in O(1) time during Two Sum in BST?

A HashSet

B Stack

C Linked List

D Queue

HashSet provides O(1) average time complexity for lookups.

What is the main advantage of using two pointers on a sorted list of BST values for Two Sum?

A It uses extra space for recursion

B It sorts the list again

C It finds pairs in O(n) time without extra space

D It only works for balanced BSTs

Two pointers scan from both ends to find pairs efficiently in O(n) time.

If the BST has n nodes, what is the space complexity of storing all node values in a list for Two Sum?

A O(log n)

B O(n)

C O(1)

D O(n log n)

Storing all node values requires O(n) space.

Which approach avoids storing all node values explicitly when solving Two Sum in BST?

A Using a HashSet during DFS traversal

B Sorting the BST nodes

C Using a queue for level order traversal

D Using recursion without extra data structures

HashSet during DFS checks complements on the fly without storing all nodes.

Explain how to solve the Two Sum problem in a BST using a HashSet during traversal.

Think about checking pairs while visiting nodes once.

Describe the two-pointer technique to find two numbers that sum to a target in a BST.

Use sorted order property of BST.

Practice

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

easy

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

B. Find the maximum value in the BST.

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, count nodes, or check balance.

3. Final Answer:

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

4. Quick Check:

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

Hint: Look for two values adding to target in BST nodes [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 of the following C# method signatures is correct for a function that checks if two nodes in a BST sum to a target value?

easy

A. string FindTarget(TreeNode root, int target)

B. bool FindTarget(TreeNode root, int target)

C. void FindTarget(TreeNode root, int target)

D. int FindTarget(TreeNode root, int target)

Solution

1. Step 1: Understand the expected return type

   The function should return true or false depending on whether two nodes sum to the target, so return type must be bool.

2. Step 2: Check method signature options

   bool FindTarget(TreeNode root, int target) returns bool, matching the requirement. Others return int, void, or string, which are incorrect for this purpose.

3. Final Answer:

   bool FindTarget(TreeNode root, int target) -> Option B

4. Quick Check:

   Return type bool for true/false check [OK]

Hint: Function returns true/false, so use bool return type [OK]

Common Mistakes:

• Using void return type and expecting a result
• Returning int or string instead of bool
• Incorrect parameter types

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

      5
     / \
    3   6
   / \   \
  2   4   7

medium

A. runtime error

B. false

C. null

D. true

Solution

1. Step 1: Identify pairs that sum to target 9

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

2. Step 2: Confirm function behavior

   Since at least one pair (e.g., 2 and 7) sums to 9, the function should return true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Pair exists with sum 9 = true [OK]

Hint: Check all node pairs for target sum quickly [OK]

Common Mistakes:

• Ignoring valid pairs in BST
• Assuming no pair without checking all nodes
• Confusing output with null or error

4. Consider this buggy code snippet for FindTarget in a BST:

public bool FindTarget(TreeNode root, int k) {
    HashSet<int> set = new HashSet<int>();
    return DFS(root, k, set);
}

private bool DFS(TreeNode node, int k, HashSet<int> set) {
    if (node == null) return false;
    if (set.Contains(k - node.val)) return true;
    set.Add(node.val);
    DFS(node.left, k, set);
    DFS(node.right, k, set);
    return false;
}

What is the main bug causing incorrect results?

medium

A. The base case should return true instead of false.

B. HashSet is not initialized properly.

C. DFS calls do not return their results, so always returns false.

D. The function should use a queue instead of recursion.

Solution

1. Step 1: Analyze DFS recursive calls

   The DFS calls to left and right subtrees are made but their return values are ignored.

2. Step 2: Understand impact on return value

   Because the recursive calls' results are not returned or checked, the function always returns false unless the first node matches.

3. Step 3: Correct fix

   Return the result of DFS calls combined with OR logic to propagate true upwards.

4. Final Answer:

   DFS calls do not return their results, so always returns false. -> Option C

5. Quick Check:

   Missing return in recursion causes false always [OK]

Hint: Always return recursive call results in DFS [OK]

Common Mistakes:

• Ignoring return values of recursive calls
• Incorrect base case return value
• Unnecessary data structure misuse

5. You want to find if there exist two nodes in a BST that sum to a target value. Which approach is the most efficient in terms of time and space? A) Convert BST to sorted list using inorder traversal, then use two pointers to find the sum. B) Use a HashSet during DFS to check complements without extra traversal. C) Check every pair of nodes by traversing the BST twice. D) Insert all node values into a queue and check sums by dequeuing pairs.

hard

A. Use HashSet during DFS to check complements.

B. Convert BST to sorted list and use two pointers.

C. Check every pair by traversing BST twice.

D. Use a queue to check sums by dequeuing pairs.

Solution

1. Step 1: Analyze each approach's efficiency

   Approach A uses O(n) space and O(n) time for traversal plus O(n) for two pointers. Approach B uses O(n) space but only one traversal with O(n) time. Approach C is O(n^2) time, inefficient. Approach D is unclear and inefficient.

2. Step 2: Choose the most efficient method

   Using a HashSet during DFS (Approach B) is efficient in time and space, as it checks complements on the fly without extra traversal.

3. Final Answer:

   Use HashSet during DFS to check complements. -> Option A

4. Quick Check:

   HashSet during DFS = efficient O(n) time and space [OK]

Hint: Use HashSet in one DFS pass for best efficiency [OK]

Common Mistakes:

• Choosing double traversal with two pointers over HashSet
• Trying all pairs causing O(n^2) time
• Using queues unnecessarily increasing complexity