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.

Practice

1. What does it mean for a binary tree to be balanced?

easy

A. The tree has the same number of nodes on left and right

B. All nodes have two children

C. The tree is a complete binary tree

D. The height difference between left and right subtrees of every node is at most 1

Solution

1. Step 1: Understand balance condition

   A balanced binary tree means the height difference between left and right subtrees of every node is at most 1.

2. Step 2: Compare other options

   Options A, B, and D describe other tree properties but not the balance condition.

3. Final Answer:

   The height difference between left and right subtrees of every node is at most 1 -> Option D

4. Quick Check:

   Balance means height difference ≤ 1 [OK]

Hint: Balance means subtree heights differ by no more than 1 [OK]

Common Mistakes:

• Confusing balanced with complete tree
• Thinking balance means equal nodes on both sides
• Assuming all nodes must have two children

2. Which of the following JavaScript function signatures is correct to check if a binary tree is balanced?

easy

A. function checkBalance(tree) { ... }

B. function isBalanced(root) { ... }

C. function balancedCheck(node) { ... }

D. function isBalancedTree() { ... }

Solution

1. Step 1: Identify common parameter for tree root

   The function should accept the root node as a parameter to check balance.

2. Step 2: Check function signatures

   Only function isBalanced(root) { ... } uses a clear parameter named 'root' which is standard for tree root nodes.

3. Final Answer:

   function isBalanced(root) { ... } -> Option B

4. Quick Check:

   Function needs root parameter [OK]

Hint: Function must accept root node parameter [OK]

Common Mistakes:

• Omitting root parameter
• Using unclear or no parameters
• Using function names that don't indicate balance check

3. What will be the output of the following code snippet?

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

function height(node) {
  if (!node) return 0;
  return 1 + Math.max(height(node.left), height(node.right));
}

function isBalanced(root) {
  if (!root) return true;
  let leftHeight = height(root.left);
  let rightHeight = height(root.right);
  if (Math.abs(leftHeight - rightHeight) > 1) return false;
  return isBalanced(root.left) && isBalanced(root.right);
}

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.left.left = new Node(5);

console.log(isBalanced(root));

medium

A. false

B. true

C. undefined

D. Throws error

Solution

1. Step 1: Calculate heights of left and right subtrees

   Left subtree height is 3 (nodes 2 -> 4 -> 5), right subtree height is 1 (node 3).

2. Step 2: Check balance condition at root

   Height difference is 3 - 1 = 2, which is greater than 1, so tree is not balanced.

3. Final Answer:

   false -> Option A

4. Quick Check:

   Height difference > 1 means not balanced [OK]

Hint: Check height difference > 1 returns false [OK]

Common Mistakes:

• Ignoring deeper subtree height
• Assuming tree is balanced if root has two children
• Confusing height with number of nodes

4. Find the bug in this code that checks if a binary tree is balanced:

function isBalanced(root) {
  if (!root) return true;
  let leftHeight = height(root.left);
  let rightHeight = height(root.right);
  if (Math.abs(leftHeight - rightHeight) > 1) return false;
  isBalanced(root.left);
  isBalanced(root.right);
  return true;
}

medium

A. Height function is not defined

B. Incorrect base case for empty tree

C. Missing return statements for recursive calls

D. Using Math.abs incorrectly

Solution

1. Step 1: Analyze recursive calls

   The calls to isBalanced(root.left) and isBalanced(root.right) are made but their results are not returned or checked.

2. Step 2: Identify missing return usage

   Without returning these recursive calls, the function always returns true after checking root, ignoring subtree balance.

3. Final Answer:

   Missing return statements for recursive calls -> Option C

4. Quick Check:

   Recursive calls must return their results [OK]

Hint: Always return recursive call results [OK]

Common Mistakes:

• Not returning recursive call results
• Assuming base case covers all
• Ignoring subtree balance checks

5. Given a binary tree, which approach is the most efficient to check if it is balanced without recalculating heights multiple times?

hard

A. Use a bottom-up recursion that returns height and balance status together

B. Check balance only at the root node

C. Traverse the tree level by level and count nodes

D. Calculate height for each node separately and check balance

Solution

1. Step 1: Understand inefficiency of separate height calculations

   Calculating height separately for each node causes repeated work, increasing time complexity.

2. Step 2: Use bottom-up recursion

   Returning height and balance status together in one recursion avoids repeated height calculations, improving efficiency.

3. Final Answer:

   Use a bottom-up recursion that returns height and balance status together -> Option A

4. Quick Check:

   Bottom-up recursion avoids repeated height calls [OK]

Hint: Combine height and balance in one recursion [OK]

Common Mistakes:

• Recalculating heights multiple times
• Checking balance only at root
• Using level order traversal for balance