Why Two Sum in BST in DSA C#?

The Big Idea

What if you could find two numbers adding to your target in a tree without checking every pair?

The Scenario

Imagine you have a big family photo album sorted by year. You want to find two photos whose years add up to a special number, but you only have the album and no index or notes.

The Problem

Looking through every possible pair of photos to find the right years takes a very long time and is easy to mess up. You might miss pairs or check the same pairs again and again.

The Solution

Using the Two Sum method in a Binary Search Tree (BST), you can quickly find if two numbers add up to the target by smartly walking through the tree from both ends, saving time and effort.

Before vs After

✗ Before

foreach (var x in list) {
  foreach (var y in list) {
    if (x + y == target) return true;
  }
}
return false;

✓ After

bool FindTwoSumInBST(TreeNode root, int target) {
  // Use two pointers from BST start and end
  // Move pointers based on sum comparison
  // Return true if found
  return false;
}

What It Enables

This method lets you find pairs adding to a target in a sorted tree quickly without checking every pair.

Real Life Example

Finding two expenses in a sorted list of transactions that add up to a budget limit, without checking every possible pair.

Key Takeaways

Manual pair checking is slow and error-prone.

Two Sum in BST uses the tree's order to find pairs efficiently.

This saves time and reduces mistakes in searching pairs.

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