Check if Binary Tree is Balanced in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a node is null.

DSA Javascript

if (node [1] null) return true;

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

A!==

B===

C<=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '===' which can cause unexpected type coercion.

Using '!=' or '!==' incorrectly.

We check if node is exactly equal to null to handle empty nodes.

2fill in blank

medium

Complete the code to calculate the height of a node recursively.

DSA Javascript

function height(node) {
  if (node === null) return 0;
  return 1 + Math.max(height(node.left), [1]);
}

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

Aheight(node.right)

Bheight(node.parent)

Cheight(node)

Dheight(node.left)

Attempts:

3 left

💡 Hint

Common Mistakes

Calling height on node.left twice instead of node.right.

Calling height on node or node.parent which causes infinite recursion or wrong result.

Height is 1 plus the maximum height of left and right subtrees.

3fill in blank

hard

Fix the error in the balance check condition.

DSA Javascript

if (Math.abs(height(node.left) - height(node.right)) [1] 1) return false;

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

A>

B<=

C===

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' which incorrectly returns false for balanced nodes.

Using '===' which only checks exact difference of 1.

A binary tree is balanced if the height difference is not more than 1, so return false if difference is greater than 1.

4fill in blank

hard

Fill both blanks to complete the recursive balanced tree check.

DSA Javascript

return isBalanced(node.[1]) && isBalanced(node.[2]);

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

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'parent' or 'child' which are not valid properties for this check.

We check if both left and right subtrees are balanced recursively.

5fill in blank

hard

Fill all three blanks to complete the full isBalanced function.

DSA Javascript

function isBalanced(node) {
  if (node [1] null) return true;
  if (Math.abs(height(node.left) - height(node.right)) [2] 1) return false;
  return isBalanced(node.[3]) && isBalanced(node.right);
}

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

A===

B>

Cleft

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' or '<=' in height difference check.

Using 'parent' or other invalid properties.

Check if node is null, then if height difference is greater than 1, then recursively check left and right subtrees.