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'

Anull

B0

Ctrue

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of null causes errors because nodes are objects.

Using true or false is incorrect because root is a node, not a boolean.

We check if the root is null to stop recursion when the node is not found.

2fill in blank

medium

Complete the code to move left or right in the BST during deletion.

DSA C#

if (key < root.val) root.left = DeleteNode(root.left, [1]);
else if (key > root.val) root.right = DeleteNode(root.right, [1]);

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

Akey

Broot.left

Croot.val

Droot.right

Attempts:

3 left

💡 Hint

Common Mistakes

Passing root.val instead of key confuses the search.

Passing root.left or root.right is incorrect because those are nodes, not keys.

We pass the key to continue searching for the node to delete.

3fill in blank

hard

Fix the error in finding the minimum node in the right subtree.

DSA C#

TreeNode minNode = root.right;
while (minNode.[1] != null) minNode = minNode.left;

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

Aparent

Bright

Cleft

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Going right finds the maximum, not minimum.

Using parent or val properties is incorrect for traversal.

To find the minimum node in the right subtree, we go left until no more left child exists.

4fill in blank

hard

Fill both blanks to replace the node's value and delete the minimum node.

DSA C#

root.val = minNode.[1];
root.right = DeleteNode(root.right, minNode.[2]);

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

Aval

Bleft

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using left or right instead of val causes wrong replacement.

Using parent is invalid because it's not used here.

We replace the root's value with the minimum node's value, then delete that minimum node by its value.

5fill in blank

hard

Fill the blanks to handle deletion when node has one or no child.

DSA C#

if (root.left == null) return root.[1];
else if (root.right == null) return root.[2];
return root;

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

Aleft

Bright

Cparent

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Returning parent or val instead of child nodes causes errors.

Mixing left and right in wrong order breaks the logic.

If left child is null, return right child to replace node; if right child is null, return left child. The last return is the original root.

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 from the root.

C. The node is replaced by its left child.

D. The node is simply removed without affecting other nodes.

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: Effect on BST structure

   Since no children exist, no restructuring is needed; just remove the node.

3. Final Answer:

   The node is simply removed without affecting other nodes. -> Option D

4. Quick Check:

   Leaf node deletion = simple removal [OK]

Hint: Leaf nodes delete by direct removal only [OK]

Common Mistakes:

• Thinking the tree restructures after leaf deletion
• Replacing leaf node with a child that doesn't exist
• Confusing leaf node deletion with internal node deletion

2. Which of the following is the correct C# syntax to delete a node with value 10 from a BST using a method named DeleteNode?

easy

A. root.DeleteNode(10);

B. root = DeleteNode(root, 10);

C. DeleteNode(root, 10);

D. DeleteNode(10, root);

Solution

1. Step 1: Understand method signature

   Usually, BST delete methods return the new root after deletion, so assignment is needed.

2. Step 2: Correct method call syntax

   Calling DeleteNode(root, 10) returns updated root, so root = DeleteNode(root, 10); is correct.

3. Final Answer:

   root = DeleteNode(root, 10); -> Option B

4. Quick Check:

   Assign returned root after deletion [OK]

Hint: Assign returned root after delete call [OK]

Common Mistakes:

• Not assigning the returned root after deletion
• Calling delete as instance method on root node
• Passing arguments in wrong order

3. Given the BST below and the C# code to delete node 15, what is the in-order traversal output after deletion?

      20
     /  \
   10    30
     
     15

root = DeleteNode(root, 15);

medium

A. [10, 20, 30]

B. [10, 15, 20, 30]

C. [20, 30]

D. [10, 20, 15, 30]

Solution

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

   Node 15 is a leaf node (no children) under 10.

2. Step 2: Delete node 15 and perform in-order traversal

   After deleting 15, in-order traversal visits nodes in ascending order: 10, 20, 30.

3. Final Answer:

   [10, 20, 30] -> Option A

4. Quick Check:

   In-order after deleting leaf = sorted without deleted node [OK]

Hint: In-order skips deleted leaf node [OK]

Common Mistakes:

• Including deleted node in traversal output
• Misordering nodes in traversal
• Confusing pre-order or post-order with in-order

4. Consider this C# method snippet for deleting a node from a BST:

Node DeleteNode(Node root, int key) {
  if (root == null) 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 == null) return root.right;
    else if (root.right == null) return root.left;
    root.val = MinValue(root.right);
    root.right = DeleteNode(root.right, root.val);
  }
  return root;
}

What is the bug in this code?

medium

A. The method does not update root.left or root.right after recursive calls.

B. The method does not handle the case when both children exist properly.

C. The method does not handle deleting leaf nodes correctly.

D. The method does not return the updated root after deletion.

Solution

1. Step 1: Analyze deletion when both children exist

   The method correctly replaces root's value with the minimum in the right subtree and deletes that minimum node recursively.

2. Step 2: Check correctness of recursive updates

   The method updates root.left and root.right after recursive calls, so this is handled properly.

3. Step 3: Identify the actual bug

   There is no bug in handling both children case or returning root; the code is correct.

4. Final Answer:

   None of the provided options correctly identify a bug. -> Option A

Hint: Check both children case carefully [OK]

Common Mistakes:

• Thinking leaf deletion is buggy here
• Missing recursive update of child pointers
• Assuming method forgets to return root

5. You have a BST with nodes: 50, 30, 70, 20, 40, 60, 80. You delete node 50 (the root). Which node replaces the root to maintain BST properties?

hard

A. Node 30 (left child of root)

B. Node 40 (maximum in left subtree)

C. Node 60 (minimum in right subtree)

D. Node 70 (right child of root)

Solution

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

   Node 50 is root with two children: left subtree (30, 20, 40) and right subtree (70, 60, 80).

2. Step 2: Find replacement node for root deletion

   Standard BST deletion replaces node with minimum node in right subtree, which is 60.

3. Final Answer:

   Node 60 (minimum in right subtree) -> Option C

4. Quick Check:

   Root replaced by min right subtree node [OK]

Hint: Replace deleted node with min right subtree node [OK]

Common Mistakes:

• Replacing with max left subtree instead of min right
• Replacing with immediate child without checking subtree
• Confusing replacement rules for BST deletion