Count Total Nodes in Binary Tree in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the total number of nodes in the binary tree.

DSA Javascript

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

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

A0

Bnull

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 1 instead of 0 for null nodes.

Returning null which causes errors in addition.

The base case returns 0 when the node is null, meaning no nodes here.

2fill in blank

medium

Complete the code to check if the current node is null.

DSA Javascript

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

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

A!==

B===

C!=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using == which allows type coercion.

Using != or !== which check inequality.

Use strict equality (===) to check if root is exactly null.

3fill in blank

hard

Fix the error in the recursive call to count nodes in the left subtree.

DSA Javascript

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

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

Aleftt

Bnode

Cright

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Typo in 'left' as 'leftt'.

Using 'right' instead of 'left'.

The left subtree is accessed by root.left, not root.leftt or others.

4fill in blank

hard

Fill both blanks to complete the function that counts nodes recursively.

DSA Javascript

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

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

A===

B==

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using loose equality (==) instead of strict (===).

Returning 1 instead of 0 for null nodes.

Use strict equality to check null and return 0 for no nodes.

5fill in blank

hard

Fill all three blanks to create a concise recursive count of nodes.

DSA Javascript

const countNodes = (root) => {
  if (root [1] null) return [2];
  return 1 + countNodes(root.[3]) + countNodes(root.right);
};

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

A===

B0

Cleft

D!==

Attempts:

3 left

💡 Hint

Common Mistakes

Using !== instead of === for null check.

Returning 1 instead of 0 for null.

Typo in 'left' property.

Check if root is null with ===, return 0, and recurse on root.left.