BST Delete Operation in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What are the three main cases to consider when deleting a node in a Binary Search Tree (BST)?

1. Node to delete is a leaf (no children).
2. Node to delete has one child.
3. Node to delete has two children.

intermediate

In BST deletion, how do you handle deleting a node with two children?

Replace the node's value with its inorder successor (smallest value in right subtree) or inorder predecessor (largest value in left subtree), then delete that successor/predecessor node.

beginner

What is the inorder successor in a BST?

The node with the smallest value in the right subtree of a given node.

intermediate

Why do we replace a node with its inorder successor or predecessor when deleting a node with two children in a BST?

Because it maintains the BST property by ensuring all left subtree nodes are smaller and right subtree nodes are larger than the replacement node.

beginner

What happens if you delete a leaf node in a BST?

You simply remove the leaf node by setting its parent's pointer to null.

Which of the following is NOT a case to consider when deleting a node in a BST?

A Node has no children

B Node has three children

C Node has two children

D Node has one child

BST nodes can have at most two children, so 'Node has three children' is not a valid case.

When deleting a node with two children, what is the first step?

A Replace the node's value with its inorder successor or predecessor

B Remove the node immediately

C Replace the node with its left child

D Replace the node with its right child

Replacing the node's value with its inorder successor or predecessor maintains BST properties before deleting the successor/predecessor node.

What is the inorder successor of a node in a BST?

A Smallest node in the right subtree

B Parent node

C Largest node in the left subtree

D Any leaf node

The inorder successor is the smallest node in the right subtree.

How do you delete a leaf node in a BST?

A Replace it with its child

B Replace it with its inorder successor

C Set its parent's pointer to null

D Replace it with its inorder predecessor

A leaf node has no children, so you remove it by setting its parent's pointer to null.

Why is the inorder successor used in deletion instead of any random node?

A It is always a leaf node

B It has two children

C It is the root node

D It maintains the BST ordering property

Using the inorder successor maintains the BST property because it is the next higher value after the node to delete.

Explain the steps to delete a node with two children in a BST.

Think about how to keep the tree ordered after deletion.

Describe how deleting a leaf node differs from deleting a node with one child in a BST.

Consider what happens to the parent's pointer.

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