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.