BST Delete Operation in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current node is null before deletion.

DSA C++

if (root == [1]) {
    return root;
}

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

Afalse

BNULL

C0

Dnullptr

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or false instead of nullptr.

Using NULL which is legacy in modern C++.

In C++, the correct way to check for a null pointer is nullptr.

2fill in blank

medium

Complete the code to find the minimum value node in the right subtree.

DSA C++

Node* minValueNode(Node* node) {
    Node* current = node;
    while (current->[1] != nullptr) {
        current = current->left;
    }
    return current;
}

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

Aleft

Bright

Cparent

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using right pointer instead of left.

Using parent or next which are not standard BST pointers.

To find the minimum value node, we keep going left until we reach a null pointer.

3fill in blank

hard

Fix the error in the recursive delete call for the right subtree.

DSA C++

root->right = deleteNode(root->right, [1]);

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

Aroot->right->val

Broot->val

Ckey

Droot->left->val

Attempts:

3 left

💡 Hint

Common Mistakes

Passing root->val instead of key.

Passing child node values instead of key.

The key to delete is passed down recursively, not the current node's value.

4fill in blank

hard

Fill both blanks to replace the node's value with the inorder successor and delete it.

DSA C++

Node* temp = [1](root->right);
root->val = temp->[2];
root->right = deleteNode(root->right, temp->val);

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

AminValueNode

BmaxValueNode

Cval

Dkey

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxValueNode instead of minValueNode.

Replacing with key instead of val.

The inorder successor is the minimum value node in the right subtree, and we replace the current node's value with it.

5fill in blank

hard

Fill all three blanks to handle the case when the node has only one child.

DSA C++

if (root->left == nullptr) {
    Node* temp = root->[1];
    delete root;
    return [2];
} else if (root->right == nullptr) {
    Node* temp = root->[3];
    delete root;
    return temp;
}

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

Aright

Btemp

Cleft

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Returning root instead of temp.

Mixing left and right pointers.

If the left child is null, we return the right child and vice versa. We store the child in temp, delete the current node, and return temp.