Two Sum in BST in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Two Sum in BST check

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

DSA C#

using System;
using System.Collections.Generic;

public class TreeNode {
    public int val;
    public TreeNode left, right;
    public TreeNode(int x) { val = x; }
}

public class Solution {
    public bool FindTarget(TreeNode root, int k) {
        var 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);
        return DFS(node.left, k, set) || DFS(node.right, k, set);
    }
}

public class Program {
    public static void Main() {
        var 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);

        var sol = new Solution();
        Console.WriteLine(sol.FindTarget(root, 9));
    }
}

A True

B False

C Compilation error

D Runtime error

💡 Hint

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

The nodes with values 2 and 7 sum to 9, so the output is True.

❓ Predict Output

intermediate

Output when no two nodes sum to target

What is the output of this C# code checking for two nodes summing to 28 in the BST?

DSA C#

using System;
using System.Collections.Generic;

public class TreeNode {
    public int val;
    public TreeNode left, right;
    public TreeNode(int x) { val = x; }
}

public class Solution {
    public bool FindTarget(TreeNode root, int k) {
        var 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);
        return DFS(node.left, k, set) || DFS(node.right, k, set);
    }
}

public class Program {
    public static void Main() {
        var 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);

        var sol = new Solution();
        Console.WriteLine(sol.FindTarget(root, 28));
    }
}

A True

B False

C Compilation error

D Runtime error

💡 Hint

Check if any two nodes sum to 28.

No two nodes in the BST sum to 28, so the output is False.

🧠 Conceptual

advanced

Understanding Two Sum in BST approach

Which of the following best describes the approach used to solve the Two Sum in BST problem in the given code?

A Perform a breadth-first search and check all pairs in O(n^2) time.

B Use a priority queue to extract nodes and check sums in O(n log n) time.

C Traverse the BST and store visited node values in a set to check complements in O(n) time.

D Sort the BST nodes into an array and use two pointers to find the sum in O(n log n) time.

💡 Hint

Think about how the code uses a HashSet during traversal.

The code uses DFS traversal and a HashSet to check if the complement exists, achieving O(n) time complexity.

🔧 Debug

advanced

Identify the error in Two Sum in BST code

What error will this code produce when run?

DSA C#

using System;
using System.Collections.Generic;

public class TreeNode {
    public int val;
    public TreeNode left, right;
    public TreeNode(int x) { val = x; }
}

public class Solution {
    public bool FindTarget(TreeNode root, int k) {
        var 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);
        return DFS(node.left, k, set) && DFS(node.right, k, set);
    }
}

public class Program {
    public static void Main() {
        var 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);

        var sol = new Solution();
        Console.WriteLine(sol.FindTarget(root, 9));
    }
}

A Always returns false

B Always returns true

C Compilation error due to missing return

D Runtime NullReferenceException

💡 Hint

Check the use of logical operators in recursion.

Using && means both left and right must be true, but the search should return true if either side finds a pair. So it returns false if one side is false.

🚀 Application

expert

Count pairs with sum in BST

Given a BST, how many unique pairs of nodes have values that sum to 10? Consider the BST below: 8 / \ 3 10 / \ \ 1 6 14 / \ / 4 7 13 Choose the correct count of unique pairs.

A 3

B 5

C 4

D 2

💡 Hint

Look for pairs of nodes whose values add up to 10.

The pairs are (3,7) and (4,6). No other pairs sum to 10 in the 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