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.

intermediate

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

Because the inorder successor or predecessor maintains the BST property after replacement, ensuring the tree remains correctly ordered.

beginner

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

The leaf node is simply removed, and its parent's pointer to it is set to null.

beginner

How do you delete a node with only one child in a BST?

Replace the node with its single child by updating the parent's pointer to point to the child, then delete the node.

Which node do you replace a node with when deleting a node with two children in a BST?

A Root node

B Inorder successor or inorder predecessor

C Any leaf node

D Random node in the tree

The inorder successor or predecessor maintains the BST property after deletion.

What is the first step when deleting a node in a BST?

A Find the node to delete

B Replace the node with root

C Delete the entire tree

D Swap left and right children

You must first find the node to delete before handling its removal.

If a node to delete has no children, what do you do?

A Replace it with inorder successor

B Replace it with its child

C Do nothing

D Remove the node and set parent's pointer to null

A leaf node can be removed directly by updating the parent's pointer.

When deleting a node with one child, what happens to the child?

A It becomes the root

B It is deleted too

C It replaces the deleted node

D It is moved to the leaf

The single child replaces the deleted node to maintain the BST structure.

Which traversal helps find the inorder successor in a BST?

A Right subtree's minimum node

B Right subtree's maximum node

C Left subtree's maximum node

D Left subtree's minimum node

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

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

Think about maintaining BST order after deletion.

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

Focus on how the parent's pointer changes.

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