BST Search Operation in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of BST Search for Existing Value

What is the output of the following C++ code that searches for the value 15 in a BST?

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Node* insert(Node* root, int val) {
    if (!root) return new Node(val);
    if (val < root->data) root->left = insert(root->left, val);
    else root->right = insert(root->right, val);
    return root;
}

Node* search(Node* root, int val) {
    if (!root || root->data == val) return root;
    if (val < root->data) return search(root->left, val);
    return search(root->right, val);
}

int main() {
    Node* root = nullptr;
    root = insert(root, 10);
    root = insert(root, 5);
    root = insert(root, 15);
    root = insert(root, 3);
    root = insert(root, 7);
    Node* result = search(root, 15);
    if (result) std::cout << result->data << std::endl;
    else std::cout << "Not found" << std::endl;
    return 0;
}

A 7

B Not found

C 10

D 15

💡 Hint

Think about how the search function returns the node if found.

The search function returns the node with data 15 because it exists in the BST. The output prints the data of the found node.

❓ Predict Output

intermediate

Output of BST Search for Non-Existing Value

What is the output of the following C++ code that searches for the value 8 in a BST?

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Node* insert(Node* root, int val) {
    if (!root) return new Node(val);
    if (val < root->data) root->left = insert(root->left, val);
    else root->right = insert(root->right, val);
    return root;
}

Node* search(Node* root, int val) {
    if (!root || root->data == val) return root;
    if (val < root->data) return search(root->left, val);
    return search(root->right, val);
}

int main() {
    Node* root = nullptr;
    root = insert(root, 10);
    root = insert(root, 5);
    root = insert(root, 15);
    root = insert(root, 3);
    root = insert(root, 7);
    Node* result = search(root, 8);
    if (result) std::cout << result->data << std::endl;
    else std::cout << "Not found" << std::endl;
    return 0;
}

A 7

B Not found

C 8

D 3

💡 Hint

Check if the value 8 exists in the BST nodes.

The value 8 is not inserted in the BST, so the search returns nullptr and prints 'Not found'.

🧠 Conceptual

advanced

Time Complexity of BST Search

What is the worst-case time complexity of searching for a value in a Binary Search Tree (BST) with n nodes?

A O(n)

B O(log n)

C O(n log n)

D O(1)

💡 Hint

Consider the shape of the BST in the worst case.

In the worst case, the BST is skewed (like a linked list), so search takes O(n) time.

🔧 Debug

advanced

Identify the Bug in BST Search Function

What error will occur when running this BST search function if the value is not found?

DSA C++

Node* search(Node* root, int val) {
    if (root == nullptr) return nullptr;
    if (root->data == val) return root;
    if (val < root->data) return search(root->left, val);
    return search(root->right, val);
}

A Segmentation fault (accessing null pointer)

B Returns nullptr correctly

C Compilation error due to missing return

D Infinite recursion

💡 Hint

What happens if root is nullptr?

The function does not check if root is nullptr before accessing root->data, causing a segmentation fault if value is not found.

🚀 Application

expert

Number of Nodes Visited During BST Search

Given a BST with nodes inserted in this order: 20, 10, 30, 5, 15, 25, 35. How many nodes are visited when searching for the value 25?

A 4

B 2

C 3

D 5

💡 Hint

Count nodes visited from root to the target node.

Search path: 20 -> 30 -> 25. Three nodes visited.

Practice

1. What is the main purpose of the search operation in a Binary Search Tree (BST)?

easy

A. To find whether a given value exists in the BST

B. To insert a new node in the BST

C. To delete a node from the BST

D. To traverse all nodes in the BST

Solution

1. Step 1: Understand the purpose of search in BST

   The search operation is used to check if a specific value is present in the BST.

2. Step 2: Differentiate from other operations

   Insertion adds nodes, deletion removes nodes, traversal visits nodes. Search only checks presence.

3. Final Answer:

   To find whether a given value exists in the BST -> Option A

4. Quick Check:

   Search = Find value [OK]

Hint: Search checks presence, not insertion or deletion [OK]

Common Mistakes:

• Confusing search with insertion
• Thinking search modifies the tree
• Mixing search with traversal

2. Which of the following is the correct syntax to start a recursive search function in C++ for a BST node pointer root and integer key?

easy

A. void search(Node root, int key) { ... }

B. bool search(Node* root, int key) { ... }

C. int search(Node* root, int key) { ... }

D. Node search(Node* root, int key) { ... }

Solution

1. Step 1: Identify return type for search

   Search returns true/false if key found, so return type should be bool.

2. Step 2: Check parameter types

   Root is a pointer to Node, key is int. So signature is bool search(Node* root, int key).

3. Final Answer:

   bool search(Node* root, int key) { ... } -> Option B

4. Quick Check:

   Return bool and pointer param [OK]

Hint: Search returns bool, takes Node pointer and key [OK]

Common Mistakes:

• Using void return type for search
• Passing Node by value instead of pointer
• Returning Node instead of bool

3. Consider the following BST and the search function call search(root, 15). What will be the output?

struct Node {
  int data;
  Node* left;
  Node* right;
};

bool search(Node* root, int key) {
  if (!root) return false;
  if (root->data == key) return true;
  if (key < root->data) return search(root->left, key);
  else return search(root->right, key);
}

// BST structure:
//       10
//      /  \
//     5    20
//         /  \
//        15  30

// Call:
// bool found = search(root, 15);
// std::cout << (found ? "Found" : "Not Found");

medium

A. Found

B. Not Found

C. Compilation Error

D. Runtime Error

Solution

1. Step 1: Trace search path for key 15

   Start at root 10, 15 > 10, go right to 20. Then 15 < 20, go left to 15.

2. Step 2: Check node data at 15

   Node data matches key 15, so function returns true and prints "Found".

3. Final Answer:

   Found -> Option A

4. Quick Check:

   Key found in BST = Found [OK]

Hint: Follow BST rules left/right to find key [OK]

Common Mistakes:

• Stopping search too early
• Confusing left and right child directions
• Assuming key not found when present

4. Identify the error in the following BST search function and choose the correct fix:

bool search(Node* root, int key) {
  if (root == nullptr) return false;
  if (root->data == key) return true;
  if (key > root->data) return search(root->left, key);
  else return search(root->right, key);
}

medium

A. Remove base case for nullptr

B. Change return type to int

C. Swap left and right in recursive calls; should be key < root->data for left

D. Add print statements for debugging

Solution

1. Step 1: Analyze recursive calls logic

   For BST, if key < root->data, search left; if key > root->data, search right.

2. Step 2: Identify incorrect direction

   Code searches left when key > root->data, which is wrong. Left subtree has smaller values.

3. Step 3: Correct the recursive calls

   Swap left and right calls: if (key < root->data) search left; else search right.

4. Final Answer:

   Swap left and right in recursive calls; should be key < root->data for left -> Option C

5. Quick Check:

   BST property fixes search direction [OK]

Hint: Left subtree < node, right subtree > node [OK]

Common Mistakes:

• Mixing left and right subtree directions
• Ignoring base case
• Changing return type unnecessarily

5. Given a BST with nodes inserted in this order: 25, 15, 50, 10, 22, 35, 70. You want to search for the value 22. How many nodes will the search function visit before finding 22?

hard

A. 2

B. 5

C. 4

D. 3

Solution

1. Step 1: Build BST structure from insertion order

   Insert 25 (root), then 15 (left of 25), 50 (right of 25), 10 (left of 15), 22 (right of 15), 35 (left of 50), 70 (right of 50).

2. Step 2: Trace search path for 22

   Start at 25 (visit 1), 22 < 25 go left to 15 (visit 2), 22 > 15 go right to 22 (visit 3), found key.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Nodes visited = 3 [OK]

Hint: Count nodes visited following BST search path [OK]

Common Mistakes:

• Counting nodes not visited
• Misplacing nodes in BST
• Stopping count before finding key