Why BST Over Plain Binary Tree in DSA Typescript - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a value into a Binary Search Tree (BST).

DSA Typescript

function insert(node: TreeNode | null, value: number): TreeNode {
  if (node === null) {
    return new TreeNode(value);
  }
  if (value [1] node.value) {
    node.left = insert(node.left, value);
  } else {
    node.right = insert(node.right, value);
  }
  return node;
}

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

A<

B>

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' causes wrong placement.

In a BST, values less than the current node go to the left subtree.

2fill in blank

medium

Complete the code to search for a value in a BST.

DSA Typescript

function search(node: TreeNode | null, value: number): boolean {
  if (node === null) return false;
  if (value === node.value) return true;
  if (value [1] node.value) {
    return search(node.left, value);
  } else {
    return search(node.right, value);
  }
}

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

A>

B==

C<

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing '<' and '>' leads to wrong search path.

To search in BST, if value is less than node's value, search left subtree.

3fill in blank

hard

Fix the error in the code that checks if a tree is a BST.

DSA Typescript

function isBST(node: TreeNode | null, min: number | null = null, max: number | null = null): boolean {
  if (node === null) return true;
  if ((min !== null && node.value [1]= min) || (max !== null && node.value [2]= max)) {
    return false;
  }
  return isBST(node.left, min, node.value) && isBST(node.right, node.value, max);
}

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

A<

B<=

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' or '>' alone misses equal values violating BST rules.

Use '<=' and '>=' to ensure no duplicates violate BST property.

4fill in blank

hard

Fill both blanks to create a function that finds the minimum value in a BST.

DSA Typescript

function findMin(node: TreeNode | null): number | null {
  if (node === null) return null;
  while (node.[1] !== null) {
    node = node.[2];
  }
  return node.value;
}

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

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' finds maximum, not minimum.

The minimum value in a BST is found by going left until no more left child exists.

5fill in blank

hard

Fill all three blanks to implement in-order traversal of a BST that collects values in an array.

DSA Typescript

function inorderTraversal(node: TreeNode | null, result: number[] = []): number[] {
  if (node !== null) {
    inorderTraversal(node.[1], result);
    result.push(node.[2]);
    inorderTraversal(node.[3], result);
  }
  return result;
}

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

Aleft

Bvalue

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing order leads to incorrect traversal sequence.

In-order traversal visits left subtree, node value, then right subtree.

Practice

1. Why is a Binary Search Tree (BST) preferred over a plain Binary Tree for searching operations?

easy

A. Because BST keeps elements in sorted order allowing faster search

B. Because BST uses less memory than a plain Binary Tree

C. Because BST nodes can have more than two children

D. Because BST does not require pointers to child nodes

Solution

1. Step 1: Understand the structure of BST vs plain Binary Tree

   A BST stores nodes so that left child is smaller and right child is larger, keeping data sorted. A plain Binary Tree has no order.

2. Step 2: Analyze search efficiency

   Because BST is sorted, searching can skip half the tree at each step, making it faster than searching a plain Binary Tree which may require checking all nodes.

3. Final Answer:

   Because BST keeps elements in sorted order allowing faster search -> Option A

4. Quick Check:

   Sorted structure = faster search [OK]

Hint: BST is sorted, plain tree is not, so search is faster [OK]

Common Mistakes:

• Thinking BST uses less memory
• Confusing number of children in BST
• Assuming BST has no pointers

2. Which of the following is the correct TypeScript syntax to declare a node in a Binary Search Tree?

easy

A. class Node { value: number; left: Node | null; right: Node | null; }

B. function Node(value) { this.value = value; this.left = null; this.right = null; }

C. interface Node { value: string; left: Node; right: Node; }

D. type Node = { value: number; left?: Node; right?: Node }

Solution

1. Step 1: Identify correct TypeScript class syntax for BST node

   class Node { value: number; left: Node | null; right: Node | null; } correctly declares a class with value and nullable left and right nodes using TypeScript types.

2. Step 2: Check other options for correctness

   function Node(value) { this.value = value; this.left = null; this.right = null; } is JavaScript function style, not TypeScript class. interface Node { value: string; left: Node; right: Node; } uses string value and non-nullable children which is incorrect. type Node = { value: number; left?: Node; right?: Node } uses optional children but is a type alias, not a class.

3. Final Answer:

   class Node { value: number; left: Node | null; right: Node | null; } -> Option A

4. Quick Check:

   TypeScript class with nullable children = class Node { value: number; left: Node | null; right: Node | null; } [OK]

Hint: Look for TypeScript class with nullable left/right nodes [OK]

Common Mistakes:

• Confusing JavaScript function with TypeScript class
• Using non-nullable children causing errors
• Using wrong value types

3. Given the following BST insertion code snippet, what will be the in-order traversal output after inserting values 10, 5, 15, 3, 7?

medium

A. [5, 3, 7, 10, 15]

B. [10, 5, 15, 3, 7]

C. [15, 10, 7, 5, 3]

D. [3, 5, 7, 10, 15]

Solution

1. Step 1: Insert values into BST in given order

   Insert 10 as root, 5 to left, 15 to right, 3 to left of 5, 7 to right of 5.

2. Step 2: Perform in-order traversal (left, root, right)

   Traverse left subtree: 3, 5, 7; then root 10; then right subtree 15. Combined: [3, 5, 7, 10, 15]

3. Final Answer:

   [3, 5, 7, 10, 15] -> Option D

4. Quick Check:

   In-order traversal of BST = sorted values [OK]

Hint: In-order traversal of BST always gives sorted list [OK]

Common Mistakes:

• Confusing traversal order
• Inserting nodes incorrectly
• Mixing up left and right children

4. Identify the bug in this BST insertion function snippet in TypeScript:

function insert(root: Node | null, val: number): Node {
  if (!root) return new Node(val);
  if (val < root.value) insert(root.left, val);
  else insert(root.right, val);
  return root;
}

medium

A. Node constructor is called without new keyword

B. Missing return statements for recursive calls

C. Incorrect comparison operator for BST insertion

D. Function does not handle duplicate values

Solution

1. Step 1: Analyze recursive calls in insertion

   The recursive calls insert(root.left, val) and insert(root.right, val) do not assign the returned node back to root.left or root.right.

2. Step 2: Identify missing return assignments

   Without assigning, the tree structure does not update. Correct code should be root.left = insert(root.left, val) and root.right = insert(root.right, val).

3. Final Answer:

   Missing return statements for recursive calls -> Option B

4. Quick Check:

   Recursive insert must update child links [OK]

Hint: Always assign recursive insert results to child nodes [OK]

Common Mistakes:

• Not assigning recursive call results
• Using wrong comparison operators
• Ignoring duplicates silently

5. You have a plain Binary Tree and a BST both containing the same values. Which operation benefits the most from using the BST instead of the plain Binary Tree, and why?

hard

A. Traversing all nodes, because BST traversal is faster than plain tree traversal

B. Inserting a new value, because BST insertion is always O(1)

C. Searching for a value, because BST allows O(log n) average time search

D. Deleting a node, because BST deletion never requires rebalancing

Solution

1. Step 1: Compare searching in BST vs plain Binary Tree

   BST keeps nodes sorted, enabling binary search-like behavior with average O(log n) time. Plain Binary Tree has no order, so search is O(n).

2. Step 2: Evaluate other operations

   Insertion in BST is not always O(1), deletion may require rebalancing, and traversal visits all nodes in both trees equally.

3. Final Answer:

   Searching for a value, because BST allows O(log n) average time search -> Option C

4. Quick Check:

   BST speeds up search, not insertion or traversal [OK]

Hint: BST speeds up search, not insertion or traversal [OK]

Common Mistakes:

• Thinking insertion is always O(1)
• Assuming deletion never needs rebalancing
• Believing traversal is faster in BST