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.