BST Search Operation in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start searching from the root node.

DSA Javascript

function searchBST(root, val) {
  let current = [1];
  while (current !== null) {
    if (val === current.val) return current;
    else if (val < current.val) current = current.left;
    else current = current.right;
  }
  return null;
}

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

Anull

Bval

Ccurrent

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing current to val instead of root

Setting current to null initially

We start searching from the root node, so current should be initialized to root.

2fill in blank

medium

Complete the code to move left when the search value is smaller than current node's value.

DSA Javascript

if (val < current.val) {
  current = [1];
}

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

Aroot.left

Bcurrent.right

Ccurrent.left

Droot.right

Attempts:

3 left

💡 Hint

Common Mistakes

Going to the right child instead of left

Using root instead of current

If the value to find is less than the current node's value, we go to the left child.

3fill in blank

hard

Fix the error in the condition to check if the current node is null.

DSA Javascript

while ([1] !== null) {
  if (val === current.val) return current;
  else if (val < current.val) current = current.left;
  else current = current.right;
}

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

Aroot

Bcurrent

Cval

Dcurrent.val

Attempts:

3 left

💡 Hint

Common Mistakes

Checking val instead of current

Checking current.val which can cause errors if current is null

The loop should continue while current is not null, meaning there are still nodes to check.

4fill in blank

hard

Fill both blanks to complete the recursive search function for BST.

DSA Javascript

function searchBST(node, val) {
  if (node === null || node.val === [1]) return node;
  if (val < node.val) return searchBST(node.[2], val);
  else return searchBST(node.right, val);
}

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

Aval

Bleft

Cright

Dnode.val

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with node.val instead of val

Using right child instead of left for smaller values

The base case checks if the current node's value equals val. If not, and val is smaller, we search the left subtree.

5fill in blank

hard

Fill all three blanks to create a function that returns true if val exists in BST, false otherwise.

DSA Javascript

function existsInBST(root, val) {
  if (root === null) return false;
  if (root.val === [1]) return true;
  if (val < root.val) return existsInBST(root.[2], [3]);
  else return existsInBST(root.right, val);
}

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

Aval

Bleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Returning true without comparison

Using right child for smaller values

Check if the current node's value equals val. If not, and val is smaller, search the left subtree recursively.

Practice

1. What is the main purpose of the search operation in a Binary Search Tree (BST)?

easy

A. To insert a new value into the BST

B. To find whether a given value exists in the BST

C. To delete a value from the BST

D. To balance the BST

Solution

1. Step 1: Understand the search operation

   The search operation in a BST is used to check if a specific value is present in the tree.

2. Step 2: Differentiate from other operations

   Insertion adds new values, deletion removes values, and balancing adjusts the tree structure. Search only checks for existence.

3. Final Answer:

   To find whether a given value exists in the BST -> Option B

4. Quick Check:

   Search = Find value [OK]

Hint: Search checks presence, not insertion or deletion [OK]

Common Mistakes:

• Confusing search with insert or delete
• Thinking search changes the tree
• Assuming search balances the tree

2. Which of the following is the correct syntax to start a search function for a BST node in JavaScript?

easy

A. function searchBST(value, root) { }

B. function searchBST(value) { return root; }

C. function searchBST(root) { return value; }

D. function searchBST(root, value) { }

Solution

1. Step 1: Identify parameters needed for BST search

   The search needs the root node to start and the value to find, so parameters are (root, value).

2. Step 2: Check function signature correctness

   function searchBST(root, value) { } correctly uses (root, value). Options A, C, and D have wrong parameter order or missing parameters.

3. Final Answer:

   function searchBST(root, value) { } -> Option D

4. Quick Check:

   Correct parameters order = function searchBST(root, value) { } [OK]

Hint: Root node first, value second in search function [OK]

Common Mistakes:

• Swapping parameter order
• Missing one parameter
• Returning wrong variable immediately

3. Consider the following BST and the search code snippet:

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

const root = new Node(10);
root.left = new Node(5);
root.right = new Node(15);

function searchBST(root, val) {
  if (!root) return false;
  if (root.val === val) return true;
  if (val < root.val) return searchBST(root.left, val);
  else return searchBST(root.right, val);
}

console.log(searchBST(root, 15));

What will be the output?

medium

A. true

B. false

C. undefined

D. Error

Solution

1. Step 1: Trace the search for value 15

   Start at root (10). Since 15 > 10, move to right child (15).

2. Step 2: Check if current node value matches 15

   Node value is 15, which matches the search value, so return true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Value found = true [OK]

Hint: Compare value with node, go left or right [OK]

Common Mistakes:

• Returning false too early
• Not checking node value before recursion
• Confusing left and right child directions

4. The following code is intended to search a value in a BST but does not work correctly. What is the error?

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

medium

A. Incorrect return type

B. Missing base case for empty tree

C. The comparison directions for left and right child are swapped

D. Function parameters are wrong

Solution

1. Step 1: Analyze comparison logic for recursion

   If val > root.val, search should go right, but code goes left. This is reversed.

2. Step 2: Confirm correct child direction for BST search

   For val > root.val, search right subtree; for val < root.val, search left subtree.

3. Final Answer:

   The comparison directions for left and right child are swapped -> Option C

4. Quick Check:

   Left/right child direction swapped = The comparison directions for left and right child are swapped [OK]

Hint: Greater value goes right, smaller goes left [OK]

Common Mistakes:

• Swapping left and right child in recursion
• Forgetting base case
• Returning wrong data type

5. You have a BST with nodes: 20 (root), 10 (left child), 30 (right child), 5 (left of 10), 15 (right of 10). You want to search for value 12. Which nodes will the search visit in order?

hard

A. 20 -> 10 -> 15

B. 20 -> 30 -> 15

C. 20 -> 10 -> 5

D. 20 -> 15 -> 10

Solution

1. Step 1: Start at root node 20 and compare with 12

   Since 12 < 20, move to left child 10.

2. Step 2: Compare 12 with node 10

   12 > 10, so move to right child 15.

3. Step 3: Compare 12 with node 15

   12 < 15, so move to left child, which is null (not in tree), search ends.

4. Final Answer:

   20 -> 10 -> 15 -> Option A

5. Quick Check:

   Visited nodes = 20, 10, 15 [OK]

Hint: Go left if smaller, right if bigger, stop if null [OK]

Common Mistakes:

• Visiting wrong child nodes
• Assuming 12 is in the tree
• Skipping nodes in path