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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

function insertNode(root, value) {
  if (root === null) {
    return { val: value, left: null, right: null };
  }
  if (value [1] root.val) {
    root.left = insertNode(root.left, value);
  } else {
    root.right = insertNode(root.right, value);
  }
  return root;
}

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 of nodes.

Using '===' or '!==' causes logical errors in comparisons.

In a BST, values less than the current node go to the left subtree, so we use < to decide where to insert.

2fill in blank

medium

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

DSA Javascript

function searchBST(root, target) {
  if (root === null) return false;
  if (root.val === target) return true;
  if (target [1] root.val) {
    return searchBST(root.left, target);
  } else {
    return searchBST(root.right, target);
  }
}

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

A<

B!==

C===

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' leads to wrong subtree search.

Not handling null root causes errors.

To search in BST, if target is less than current node, search left subtree.

3fill in blank

hard

Complete the code to find the minimum value in a BST.

DSA Javascript

function findMin(root) {
  while (root.[1] !== null) {
    root = root.left;
  }
  return root.val;
}

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

Aval

Bright

Cleft

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' will not find the minimum value.

Using root.val in condition is incorrect.

The minimum value in a BST is located at the leftmost node. Therefore, we keep moving left while root.left !== null.

4fill in blank

hard

Fill both blanks to create a function that counts nodes in a BST.

DSA Javascript

function countNodes(root) {
  if (root === null) return 0;
  return 1 + countNodes(root.[1]) + countNodes(root.[2]);
}

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

Aleft

Bright

Cval

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using root.val or root.parent instead of left/right causes errors.

Not adding 1 for current node leads to wrong count.

To count all nodes, recursively count left and right subtrees using root.left and root.right.

5fill in blank

hard

Fill all three blanks to create a function that deletes a node from a BST.

DSA Javascript

function deleteNode(root, key) {
  if (root === null) return root;
  if (key [1] root.val) {
    root.left = deleteNode(root.left, key);
  } else if (key [2] root.val) {
    root.right = deleteNode(root.right, key);
  } else {
    if (root.left === null) return root.[3];
    else if (root.right === null) return root.left;
    let minNode = findMinNode(root.right);
    root.val = minNode.val;
    root.right = deleteNode(root.right, minNode.val);
  }
  return root;
}

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

A<

B>

Cright

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping < and > causes wrong traversal.

Returning root.left instead of root.right when left is null causes errors.

To delete a node, compare key with root.val using < and > to traverse left or right. If node has only right child, return root.right.

Practice

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

easy

A. BST does not require pointers to child nodes.

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

C. BST nodes can have more than two children.

D. BST allows faster search by maintaining sorted order of elements.

Solution

1. Step 1: Understand the structure difference

   A BST keeps elements in sorted order where left child is smaller and right child is larger, unlike a plain Binary Tree which has no order.

2. Step 2: Analyze search efficiency

   Because of this order, BST allows searching by comparing and moving left or right, reducing search time to O(log n) on average, while plain Binary Tree search is O(n).

3. Final Answer:

   BST allows faster search by maintaining sorted order of elements. -> Option D

4. Quick Check:

   BST sorted order = faster search [OK]

Hint: BST keeps order for quick search decisions [OK]

Common Mistakes:

• Thinking BST uses less memory than plain Binary Tree
• Confusing number of children in BST
• Assuming BST nodes don't have child pointers

2. Which of the following is the correct way to check if a node's value is less than the root's value in JavaScript BST code?

easy

A. if (node.value < root.value) {}

B. if node.value < root.value {}

C. if (node.value << root.value) {}

D. if node.value < root.value then {}

Solution

1. Step 1: Identify correct JavaScript syntax for if statement

   JavaScript requires parentheses around the condition and curly braces for the block.

2. Step 2: Check comparison operator and syntax

   The less than operator is <, and it must be inside parentheses: if (condition) {} is correct.

3. Final Answer:

   if (node.value < root.value) {} -> Option A

4. Quick Check:

   Correct if syntax = if (node.value < root.value) {} [OK]

Hint: Use parentheses and curly braces in if statements [OK]

Common Mistakes:

• Omitting parentheses around condition
• Using incorrect operators like <<
• Writing if without parentheses or braces

3. What will be the output of the following code snippet that inserts values into a BST and then performs an in-order traversal?

class Node {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

class BST {
  constructor() {
    this.root = null;
  }
  insert(value) {
    const newNode = new Node(value);
    if (!this.root) {
      this.root = newNode;
      return;
    }
    let current = this.root;
    while (true) {
      if (value < current.value) {
        if (!current.left) {
          current.left = newNode;
          break;
        }
        current = current.left;
      } else {
        if (!current.right) {
          current.right = newNode;
          break;
        }
        current = current.right;
      }
    }
  }
  inOrder(node, result = []) {
    if (node) {
      this.inOrder(node.left, result);
      result.push(node.value);
      this.inOrder(node.right, result);
    }
    return result;
  }
}

const tree = new BST();
tree.insert(10);
tree.insert(5);
tree.insert(15);
tree.insert(7);
console.log(tree.inOrder(tree.root));

medium

A. [5, 7, 10, 15]

B. [10, 5, 7, 15]

C. [15, 10, 7, 5]

D. [7, 5, 10, 15]

Solution

1. Step 1: Insert nodes into BST

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

2. Step 2: Perform in-order traversal

   In-order visits left subtree, root, then right subtree. So order is 5, 7, 10, 15.

3. Final Answer:

   [5, 7, 10, 15] -> Option A

4. Quick Check:

   In-order BST traversal = sorted values [OK]

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

Common Mistakes:

• Confusing traversal order
• Inserting nodes incorrectly
• Expecting pre-order or post-order output

4. The following code attempts to insert a value into a BST but causes an infinite loop. What is the error?

insert(value) {
  const newNode = new Node(value);
  if (!this.root) {
    this.root = newNode;
    return;
  }
  let current = this.root;
  while (true) {
    if (value < current.value) {
      if (!current.left) {
        current.left = newNode;
        break;
      } else {
        current = current.left;
      }
    } else {
      if (!current.right) {
        current.right = newNode;
        break;
      } else {
        current = current.right;
      }
    }
  }
}

medium

A. New node is not created properly.

B. Incorrect comparison operator used.

C. Missing break statement after inserting new node.

D. Root node is not initialized.

Solution

1. Step 1: Analyze loop behavior after insertion

   After assigning newNode to left or right child, loop continues without break, causing infinite loop.

2. Step 2: Identify missing break

   Adding break after insertion stops the loop and prevents infinite iteration.

3. Final Answer:

   Missing break statement after inserting new node. -> Option C

4. Quick Check:

   Break after insert stops infinite loop [OK]

Hint: Add break after insertion to exit loop [OK]

Common Mistakes:

• Forgetting break after insertion
• Changing comparison operators unnecessarily
• Assuming new node creation is faulty

5. You want to store a large set of numbers and perform frequent searches. Why would using a balanced BST be better than a plain Binary Tree or an unbalanced BST?

hard

A. Balanced BST uses less memory than unbalanced BST.

B. Balanced BST guarantees O(log n) search time, unlike unbalanced BST or plain Binary Tree.

C. Balanced BST allows nodes to have multiple parents for faster access.

D. Balanced BST does not require pointers to child nodes.

Solution

1. Step 1: Understand search time in different trees

   Plain Binary Tree search is O(n), unbalanced BST can degrade to O(n) if skewed, balanced BST keeps height low.

2. Step 2: Benefit of balanced BST

   Balanced BST maintains height around log n, ensuring search, insert, delete operations are O(log n) consistently.

3. Final Answer:

   Balanced BST guarantees O(log n) search time, unlike unbalanced BST or plain Binary Tree. -> Option B

4. Quick Check:

   Balanced BST = consistent fast search [OK]

Hint: Balance keeps tree height low for fast search [OK]

Common Mistakes:

• Thinking balanced BST uses less memory
• Believing nodes can have multiple parents
• Assuming no child pointers needed