BST balancing problem in Data Structures Theory - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Why is balancing important in a Binary Search Tree?

Consider a Binary Search Tree (BST). Why is it important to keep the tree balanced?

A To keep the height of the tree low, improving search, insert, and delete operations.

B To allow the tree to store duplicate values efficiently.

C To make sure all nodes have two children.

D To ensure the tree uses the least memory possible.

💡 Hint

Think about how the height of the tree affects the time it takes to find a value.

Balancing a BST keeps its height low, which means operations like search, insert, and delete can be done quickly, typically in O(log n) time. An unbalanced tree can degrade to a linked list with O(n) time.

📋 Factual

intermediate

What is the maximum height of a balanced BST with n nodes?

What is the maximum height of a balanced Binary Search Tree containing n nodes?

A log₂(n)

B n/2

C n log n

D n

💡 Hint

Balanced trees aim for the smallest possible height relative to the number of nodes.

A balanced BST has height proportional to log₂(n), which ensures efficient operations.

🔍 Analysis

advanced

Identify the imbalance type in this BST after insertion

After inserting a new node into a BST, the tree becomes unbalanced. The imbalance occurs when a node's left child's right subtree is taller than its left subtree. What type of imbalance is this?

A Left-Left (LL) imbalance

B Left-Right (LR) imbalance

C Right-Right (RR) imbalance

D Right-Left (RL) imbalance

💡 Hint

Think about which subtree of the left child is causing the imbalance.

When the left child's right subtree is taller, it is a Left-Right (LR) imbalance, which requires a double rotation to fix.

❓ Comparison

advanced

Compare AVL and Red-Black Trees in balancing

Which statement correctly compares AVL trees and Red-Black trees regarding balancing?

A AVL trees are less strictly balanced than Red-Black trees, so they have faster insertions.

B Red-Black trees are more strictly balanced than AVL trees, leading to faster searches.

C AVL trees are more strictly balanced than Red-Black trees, resulting in faster lookups but slower insertions and deletions.

D Both AVL and Red-Black trees have the same balancing rules and performance characteristics.

💡 Hint

Consider how strict the balancing rules are and how that affects operation speeds.

AVL trees maintain a stricter balance, which makes searches faster but insertions and deletions slower due to more rotations. Red-Black trees allow more imbalance, making insertions and deletions faster but searches slightly slower.

❓ Reasoning

expert

Determining the number of rotations needed to balance a BST

Given a BST that becomes unbalanced after inserting a node causing a Right-Left (RL) imbalance at a certain node, how many rotations are needed to restore balance?

A One single rotation

B Three rotations: two left rotations and one right rotation

C Two rotations: first right rotation on the right child, then left rotation on the unbalanced node

D Two rotations: first left rotation on the right child, then right rotation on the unbalanced node

💡 Hint

Think about the standard fix for RL imbalance in AVL trees.

For a Right-Left imbalance, the fix is a double rotation: first a left rotation on the right child, then a right rotation on the unbalanced node.

Practice

1. What is the main reason to balance a Binary Search Tree (BST)?

easy

A. To keep operations like search, insert, and delete efficient

B. To increase the number of nodes in the tree

C. To make the tree look symmetrical

D. To store duplicate values easily

Solution

1. Step 1: Understand BST operations

   BST operations like search, insert, and delete depend on tree height for speed.

2. Step 2: Effect of balancing

   Balancing keeps the tree height minimal, making these operations faster.

3. Final Answer:

   To keep operations like search, insert, and delete efficient -> Option A

4. Quick Check:

   Balanced BST = Efficient operations [OK]

Hint: Balanced BST means faster search and updates [OK]

Common Mistakes:

• Thinking balancing increases nodes
• Believing balancing is just for looks
• Confusing balancing with allowing duplicates

2. Which of the following is a correct step in balancing a BST?

easy

A. Get sorted keys and rebuild tree with middle key as root

B. Delete all leaf nodes

C. Insert nodes in ascending order

D. Swap left and right children of all nodes

Solution

1. Step 1: Recall balancing method

   Balancing involves sorting keys and rebuilding the tree.

2. Step 2: Middle key as root

   Choosing the middle key as root ensures minimal height and balance.

3. Final Answer:

   Get sorted keys and rebuild tree with middle key as root -> Option A

4. Quick Check:

   Middle key root = Balanced BST [OK]

Hint: Middle key root rebuilds balanced BST [OK]

Common Mistakes:

• Inserting nodes in sorted order causes unbalance
• Deleting leaf nodes does not balance tree
• Swapping children does not guarantee balance

3. Given a BST with nodes inserted in this order: 10, 5, 1, 7, 40, 50, what is the height of the tree after balancing it?

medium

A. 2

B. 5

C. 4

D. 3

Solution

1. Step 1: List nodes in sorted order

   Sorted keys: 1, 5, 7, 10, 40, 50.

2. Step 2: Build balanced BST

   Middle key is 10 (root), left subtree with 1,5,7, right subtree with 40,50.

3. Step 3: Calculate height

   Height is 3 (longest path from root to leaf has 3 edges): root(10), children(5,40), grandchildren(1,7,50).

4. Final Answer:

   3 -> Option D

5. Quick Check:

   Balanced BST height = 3 [OK]

Hint: Balanced BST height ~ log2(n) [OK]

Common Mistakes:

• Counting unbalanced height
• Choosing wrong middle key
• Miscounting tree levels

4. You tried to balance a BST by just swapping left and right children of every node. What is the main problem with this approach?

medium

A. It sorts the keys incorrectly

B. It deletes all leaf nodes accidentally

C. It may create an unbalanced tree or violate BST property

D. It duplicates nodes in the tree

Solution

1. Step 1: Understand swapping effect

   Swapping children flips the tree but does not reorder keys.

2. Step 2: Check BST property

   BST requires left child keys < node < right child keys; swapping breaks this.

3. Final Answer:

   It may create an unbalanced tree or violate BST property -> Option C

4. Quick Check:

   Swapping children ≠ balancing BST [OK]

Hint: Swapping children breaks BST order, not balance [OK]

Common Mistakes:

• Thinking swapping sorts keys
• Assuming swapping deletes nodes
• Believing swapping duplicates nodes

5. You have a BST with keys [3, 8, 10, 15, 20, 25, 30] inserted in that order. To balance it, you extract the sorted keys and rebuild the tree. Which key should be the root of the balanced BST?

medium

A. 10

B. 15

C. 20

D. 8

Solution

1. Step 1: Sort keys

   Keys are already sorted: 3, 8, 10, 15, 20, 25, 30.

2. Step 2: Find middle key

   Middle key is the 4th element (index 3): 15.

3. Step 3: Use middle key as root

   Root = 15 ensures balanced left and right subtrees.

4. Final Answer:

   15 -> Option B

5. Quick Check:

   Middle key root = 15 [OK]

Hint: Middle element of sorted keys is balanced root [OK]

Common Mistakes:

• Choosing first or last key as root
• Picking a key not in the middle
• Ignoring sorted order