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DSA C#programming~5 mins

Two Sum in BST in DSA C# - Time & Space Complexity

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Time Complexity: Two Sum in BST
O(n)
Understanding Time Complexity

We want to know how the time needed to find two numbers that add up to a target grows as the tree gets bigger.

How does the search time change when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


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);
}
    

This code searches the tree to find if two values add up to the target by checking each node once and storing seen values.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Recursive traversal of each node in the tree.
  • How many times: Each node is visited exactly once during the depth-first search.
How Execution Grows With Input

As the number of nodes grows, the function visits each node once, so the work grows directly with the number of nodes.

Input Size (n)Approx. Operations
10About 10 visits and checks
100About 100 visits and checks
1000About 1000 visits and checks

Pattern observation: The operations increase in a straight line as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the two sum grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "Because the tree is a BST, we can find the pair in less than linear time by skipping nodes."

[OK] Correct: The algorithm must check each node to be sure, so it still visits all nodes once, making it linear time.

Interview Connect

Understanding this helps you explain how searching in trees works and shows you can analyze recursive code clearly.

Self-Check

"What if we used an in-order traversal to get a sorted list first, then used two pointers to find the sum? How would the time complexity change?"

Practice

(1/5)
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
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