BST Insert Operation in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node with the given value.

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 == nullptr) {
        return new [1](val);
    }
    // rest of insert logic
    return root;
}

Drag options to blanks, or click blank then click option'

ANode*

BTree

CBST

DNode

Attempts:

3 left

💡 Hint

Common Mistakes

Using pointer type 'Node*' instead of the struct name 'Node'.

Using an undefined type like 'Tree' or 'BST'.

We create a new node by calling the Node constructor with the value.

2fill in blank

medium

Complete the code to insert the value into the left subtree if it is smaller than the root's data.

DSA C++

if (val < root->data) {
    root->left = [1](root->left, val);
}

Drag options to blanks, or click blank then click option'

Ainsert

BnewNode

CcreateNode

DaddNode

Attempts:

3 left

💡 Hint

Common Mistakes

Using a function name that does not exist like 'newNode' or 'createNode'.

Not calling the function recursively.

We recursively call insert on the left child to place the new value correctly.

3fill in blank

hard

Fix the error in the code to insert the value into the right subtree if it is greater than the root's data.

DSA C++

if (val > root->data) {
    root->right = [1](root->right, val);
}

Drag options to blanks, or click blank then click option'

Ainsert

BinsertNode

CaddRight

DnewNode

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect function names like 'insertNode' or 'addRight'.

Not calling the function recursively.

The correct function to call recursively is 'insert' to maintain BST properties.

4fill in blank

hard

Fill both blanks to complete the insert function that returns the root after insertion.

DSA C++

Node* insert(Node* root, int val) {
    if (root == nullptr) {
        return new [1](val);
    }
    if (val < root->data) {
        root->left = [2](root->left, val);
    } else if (val > root->data) {
        root->right = insert(root->right, val);
    }
    return root;
}

Drag options to blanks, or click blank then click option'

ANode

Binsert

CNode*

DaddNode

Attempts:

3 left

💡 Hint

Common Mistakes

Using pointer type 'Node*' instead of 'Node' for new node creation.

Using wrong function names for recursion.

We create a new Node and recursively call insert on the left subtree.

5fill in blank

hard

Fill all three blanks to complete the insert function with handling for equal values (ignore duplicates).

DSA C++

Node* insert(Node* root, int val) {
    if (root == nullptr) {
        return new [1](val);
    }
    if (val < root->data) {
        root->left = [2](root->left, val);
    } else if (val > root->data) {
        root->right = [3](root->right, val);
    } else {
        // Duplicate value, do nothing
        return root;
    }
    return root;
}

Drag options to blanks, or click blank then click option'

ANode

Binsert

DNode*

Attempts:

3 left

💡 Hint

Common Mistakes

Using pointer type 'Node*' instead of 'Node' for new node creation.

Not handling duplicates properly.

Using wrong function names for recursion.

We create a new Node and recursively call insert on left or right subtree. Equal values are ignored.

Practice

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

easy

A. To delete a node from the BST

B. To find the height of the BST

C. To add a new node while maintaining BST properties

D. To traverse the BST in order

Solution

1. Step 1: Understand BST insert operation

   The insert operation adds a new node to the BST while keeping the order property intact: left child < parent < right child.

2. Step 2: Differentiate from other operations

   Deleting nodes, finding height, or traversing are different operations and do not describe insertion.

3. Final Answer:

   To add a new node while maintaining BST properties -> Option C

4. Quick Check:

   Insert = Add node maintaining order [OK]

Hint: Insert means add node keeping BST order intact [OK]

Common Mistakes:

• Confusing insert with delete operation
• Thinking insert changes tree height only
• Mixing insert with traversal methods

2. Which of the following is the correct syntax to insert a value val into a BST using a recursive function in C++?

easy

A. root = insert(root, val);

B. insert(root, val);

C. root.insert(val);

D. insert(val, root);

Solution

1. Step 1: Understand recursive insert signature

   Recursive insert usually returns the new root after insertion, so assignment like root = insert(root, val); is needed.

2. Step 2: Check other options

   Calling insert(root, val); without assignment ignores updated root. root.insert(val); assumes method on node object which is uncommon. insert(val, root); has wrong parameter order.

3. Final Answer:

   root = insert(root, val); -> Option A

4. Quick Check:

   Assign returned root after insert [OK]

Hint: Assign root = insert(root, val) for recursive insertion [OK]

Common Mistakes:

• Not assigning the returned root after insertion
• Using wrong parameter order in insert call
• Assuming insert is a method of node object

3. Consider the following C++ code inserting values into an empty BST:

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;
}

int main() {
  Node* root = nullptr;
  root = insert(root, 10);
  root = insert(root, 5);
  root = insert(root, 15);
  root = insert(root, 12);
  return 0;
}

What is the structure of the BST after these insertions?

medium

A. 10 -> 5 -> 12 (left), 10 -> 15 -> null (right subtree)

B. 10 -> 15 -> null (left), 10 -> 5 -> 12 (right subtree)

C. 10 -> 12 -> 5 (left), 10 -> 15 -> null (right subtree)

D. 10 -> 5 -> null (left), 10 -> 15 -> 12 (right subtree)

Solution

1. Step 1: Insert 10, 5, 15, 12 step-by-step

   Start with empty root. Insert 10 as root. Insert 5 goes left of 10. Insert 15 goes right of 10. Insert 12 goes left of 15.

2. Step 2: Visualize BST structure

   Root 10 has left child 5 (no children), right child 15 with left child 12 (no right child).

3. Final Answer:

   10 -> 5 -> null (left), 10 -> 15 -> 12 (right subtree) -> Option D

4. Quick Check:

   Insert order builds left 5, right 15 with left 12 [OK]

Hint: Insert smaller left, larger right; 12 is left child of 15 [OK]

Common Mistakes:

• Placing 12 as right child of 15 instead of left
• Swapping left and right children of 10
• Ignoring null children in output

4. The following code attempts to insert a value into a BST but causes a runtime error:

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

What is the error and how to fix it?

medium

A. Missing return after creating new node; add return newNode;

B. Incorrect comparison operator; use <= instead of <

C. Should not assign to root->left or root->right

D. Node constructor is missing initialization

Solution

1. Step 1: Identify missing return in base case

   When root is nullptr, new node is created but not returned, so function continues and dereferences nullptr causing runtime error.

2. Step 2: Fix by returning new node immediately

   Add return newNode; inside the if block to return the newly created node and stop further execution.

3. Final Answer:

   Missing return after creating new node; add return newNode; -> Option A

4. Quick Check:

   Return new node on nullptr base case [OK]

Hint: Always return new node immediately when root is nullptr [OK]

Common Mistakes:

• Not returning new node causes nullptr dereference
• Changing comparison operator unnecessarily
• Misunderstanding pointer assignments

5. You have a BST with nodes 20, 10, 30, 25, 40. You want to insert 15 and 35. After insertion, which of the following is the correct in-order traversal of the BST?

hard

A. [15, 10, 20, 25, 30, 35, 40]

B. [10, 15, 20, 25, 30, 35, 40]

C. [10, 20, 15, 25, 30, 35, 40]

D. [10, 15, 20, 30, 25, 35, 40]

Solution

1. Step 1: Insert 15 and 35 into BST

   Insert 15: less than 20, greater than 10, goes right of 10. Insert 35: less than 40, greater than 30, goes right of 30.

2. Step 2: Perform in-order traversal

   In-order traversal visits nodes in ascending order: left subtree, root, right subtree. The order is 10, 15, 20, 25, 30, 35, 40.

3. Final Answer:

   [10, 15, 20, 25, 30, 35, 40] -> Option B

4. Quick Check:

   In-order traversal = sorted nodes [OK]

Hint: In-order traversal of BST is sorted list [OK]

Common Mistakes:

• Placing 15 or 35 in wrong subtree
• Confusing traversal order
• Listing nodes unsorted in output