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.

Practice

1. What happens when you delete a leaf node (a node with no children) from a Binary Search Tree (BST)?

easy

A. The node is replaced by its right child.

B. The entire tree is restructured to maintain BST properties.

C. The node is simply removed and its parent's pointer is set to null.

D. The node is replaced by its left child.

Solution

1. Step 1: Understand leaf node deletion

   A leaf node has no children, so removing it does not affect other nodes.

2. Step 2: Update parent's pointer

   The parent's pointer to this node is set to null, effectively removing the node.

3. Final Answer:

   The node is simply removed and its parent's pointer is set to null. -> Option C

4. Quick Check:

   Leaf node deletion = remove node and nullify parent's pointer [OK]

Hint: Leaf node deletion just removes the node, no restructuring needed [OK]

Common Mistakes:

• Thinking the tree needs full restructuring
• Replacing leaf node with children when none exist
• Confusing leaf node deletion with internal node deletion

2. Which of the following is the correct syntax to delete a node with value key from a BST using a recursive function named deleteNode in C++?

easy

A. root.deleteNode(key);

B. deleteNode(root, key);

C. deleteNode(key, root);

D. root = deleteNode(root, key);

Solution

1. Step 1: Understand function signature

   The function returns the new root after deletion, so assignment is needed.

2. Step 2: Correct function call

   Calling root = deleteNode(root, key); updates the root pointer correctly.

3. Final Answer:

   root = deleteNode(root, key); -> Option D

4. Quick Check:

   Assign returned root after deletion [OK]

Hint: Always assign root = deleteNode(root, key) to update tree [OK]

Common Mistakes:

• Not assigning the returned root to update the tree
• Swapping argument order incorrectly
• Using method call syntax on root pointer

3. Consider the BST below and the code snippet that deletes the node with value 10:

    15
   /  \
  10  20
 /    / \
8    17 25

root = deleteNode(root, 10); What will be the in-order traversal of the BST after deletion?

medium

A. [8, 15, 10, 17, 20, 25]

B. [8, 15, 17, 20, 25]

C. [8, 15, 17, 20]

D. [8, 10, 15, 17, 20, 25]

Solution

1. Step 1: Identify node to delete and its children

   Node 10 has one child (8). Deleting 10 replaces it with 8.

2. Step 2: Perform in-order traversal after deletion

   In-order traversal visits nodes in ascending order: 8, 15, 17, 20, 25.

3. Final Answer:

   [8, 15, 17, 20, 25] -> Option B

4. Quick Check:

   In-order after deleting 10 = [8, 15, 17, 20, 25] [OK]

Hint: Replace deleted node with child or successor, then traverse in-order [OK]

Common Mistakes:

• Forgetting to replace node 10 with its child 8
• Including deleted node 10 in traversal
• Misordering nodes in traversal

4. The following C++ code attempts to delete a node from a BST but has a bug. What is the bug?

Node* deleteNode(Node* root, int key) {
  if (!root) return root;
  if (key < root->val) root->left = deleteNode(root->left, key);
  else if (key > root->val) root->right = deleteNode(root->right, key);
  else {
    if (!root->left) return root->right;
    else if (!root->right) return root->left;
    Node* temp = root->left; // BUG HERE
    root->val = temp->val;
    root->left = deleteNode(root->left, temp->val);
  }
  return root;
}

medium

A. It uses the left child instead of the inorder successor (right subtree minimum) for replacement.

B. It does not handle the case when root is null.

C. It deletes the wrong key by comparing incorrectly.

D. It does not update the root pointer after deletion.

Solution

1. Step 1: Identify replacement node for deletion

   For nodes with two children, the inorder successor (minimum in right subtree) should replace the node.

2. Step 2: Spot incorrect replacement

   The code incorrectly uses the left child as replacement, which breaks BST properties.

3. Final Answer:

   It uses the left child instead of the inorder successor (right subtree minimum) for replacement. -> Option A

4. Quick Check:

   Use inorder successor (right subtree min) for two-child deletion [OK]

Hint: Replace with inorder successor (min right subtree), not left child [OK]

Common Mistakes:

• Using left child instead of inorder successor
• Not updating pointers after replacement
• Ignoring null root check

5. You have a BST with nodes: 30, 20, 40, 10, 25, 35, 50. You delete node 30 (root). Which node replaces the root to maintain BST properties?

hard

A. Node 35 (minimum node in right subtree)

B. Node 25 (maximum node in left subtree)

C. Node 40 (right child of root)

D. Node 20 (left child of root)

Solution

1. Step 1: Identify node to delete and its children

   Node 30 is root with two children: left subtree (20) and right subtree (40).

2. Step 2: Find inorder successor to replace root

   Inorder successor is the minimum node in right subtree, which is 35.

3. Step 3: Replace root with inorder successor

   Node 35 replaces 30 to maintain BST properties after deletion.

4. Final Answer:

   Node 35 (minimum node in right subtree) -> Option A

5. Quick Check:

   Root replacement = inorder successor (min right subtree) [OK]

Hint: Replace deleted root with inorder successor (min right subtree) [OK]

Common Mistakes:

• Replacing with left subtree max instead of right subtree min
• Replacing with direct child without checking subtree
• Confusing inorder predecessor and successor