BST Search Operation in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a Binary Search Tree (BST)?

A BST is a tree where each node has at most two children. The left child's value is less than the parent's value, and the right child's value is greater. This helps in fast searching.

beginner

How does the search operation work in a BST?

Start at the root. If the value equals the node's value, stop. If smaller, go left. If larger, go right. Repeat until found or no more nodes.

beginner

Why is BST search faster than searching in a list?

BST search skips half of the tree at each step by choosing left or right child based on comparison, unlike a list which checks each item one by one.

intermediate

What is the time complexity of searching in a balanced BST?

The time complexity is O(log n) because each step halves the search space in a balanced BST.

beginner

What happens if the value is not found in the BST during search?

The search reaches a null child (no node), meaning the value is not in the tree.

In a BST, if the search value is less than the current node's value, where do you go next?

A Root node

B Right child

C Parent node

D Left child

If the search value is less, you move to the left child because all left children have smaller values.

What is the best case time complexity of searching in a BST?

A O(n)

B O(log n)

C O(1)

D O(n log n)

Best case is O(1) when the root node itself contains the search value.

If a BST is not balanced, what can happen to the search time?

A It becomes O(n)

B It becomes O(1)

C It stays O(log n)

D It becomes O(n log n)

If the BST is skewed (like a linked list), search time can degrade to O(n).

What is the first step in searching a value in a BST?

A Check the left child

B Compare with the root node

C Check the right child

D Insert the value

Search always starts by comparing the value with the root node.

If the search value is greater than the current node's value, where do you move next?

A Right child

B Left child

C Parent node

D Root node

You move to the right child because all right children have greater values.

Explain the step-by-step process of searching a value in a BST.

Think about how you decide which child to visit next.

Describe why BST search is more efficient than searching in an unsorted list.

Focus on how BST structure helps reduce search steps.

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