Two Sum in BST in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the BST node is null.

DSA C#

if (root [1] null) 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 '==' causes wrong logic.

Using comparison operators like '<=' or '>' with null is invalid.

We check if the current node is null using '== null' to stop recursion.

2fill in blank

medium

Complete the code to add a value to the HashSet.

DSA C#

set.[1](root.val);

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

AContains

BClear

CAdd

DRemove

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Contains' checks existence but does not add.

Using 'Remove' deletes elements, which is incorrect here.

We add the current node's value to the set using 'Add'.

3fill in blank

hard

Fix the error in the condition to check if complement exists in the set.

DSA C#

if (set.[1](target - root.val)) return true;

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

AContains

BAdd

CRemove

DClear

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Add' instead of 'Contains' causes wrong logic.

Using 'Remove' or 'Clear' is incorrect here.

We use 'Contains' to check if the complement value exists in the set.

4fill in blank

hard

Fill both blanks to recursively search left and right subtrees.

DSA C#

return TwoSumBST(root.[1], target, set) || TwoSumBST(root.[2], target, set);

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

Aleft

Bright

Cparent

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'parent' or 'root' properties which do not exist causes errors.

Checking only one side misses possible pairs.

We recursively check left and right child nodes to find the pair.

5fill in blank

hard

Fill all three blanks to complete the TwoSumBST method signature and initial call.

DSA C#

public bool TwoSumBST(TreeNode [1], int [2]) {
    var set = new HashSet<int>();
    return TwoSumBST([3], [2], set);
}

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

Aroot

Btarget

Cthis.root

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'value' instead of 'target' confuses the target sum.

Using 'this.root' inside the method signature is invalid.

The method takes 'root' and 'target' as parameters, and calls the helper with 'root' and 'target'.

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