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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the height of the tree or -1 if unbalanced.

DSA Typescript

function checkHeight(node: TreeNode | null): number {
  if (node === null) return [1];
  // rest of code omitted for brevity
}

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

A0

B-1

Cnull

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Returning -1 for null node height causes incorrect balance checks.

When the node is null, height is 0 because empty tree has height zero.

2fill in blank

medium

Complete the code to check if the left subtree is balanced.

DSA Typescript

const leftHeight = checkHeight(node.left);
if (leftHeight === [1]) return -1;

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

A0

Bnull

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for 0 instead of -1 causes missing unbalanced detection.

If left subtree is unbalanced, checkHeight returns -1, so we propagate -1 upwards.

3fill in blank

hard

Fix the error in the balance difference check.

DSA Typescript

if (Math.abs(leftHeight - rightHeight) > [1]) return -1;

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

A1

B0

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 2 causes incorrect balance detection.

The tree is unbalanced if height difference is more than 1.

4fill in blank

hard

Fill both blanks to return the height of the current node.

DSA Typescript

return Math.max([1], [2]) + 1;

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

AleftHeight

BrightHeight

Cnode.left

Dnode.right

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.left or node.right instead of their heights causes errors.

The height of current node is max of left and right subtree heights plus one.

5fill in blank

hard

Fill all three blanks to complete the isBalanced function.

DSA Typescript

function isBalanced(root: TreeNode | null): boolean {
  return checkHeight([1]) !== [2];
}

function checkHeight(node: TreeNode | null): number {
  if (node === null) return [3];
  const leftHeight = checkHeight(node.left);
  if (leftHeight === -1) return -1;
  const rightHeight = checkHeight(node.right);
  if (rightHeight === -1) return -1;
  if (Math.abs(leftHeight - rightHeight) > 1) return -1;
  return Math.max(leftHeight, rightHeight) + 1;
}

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

Aroot

B-1

C0

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Returning true if checkHeight returns -1 causes wrong result.

Returning -1 for null node height causes errors.

Call checkHeight on root, return true if result is not -1, and height of null node is 0.

Practice

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

easy

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

B. All nodes have two children

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

D. The tree is a complete binary tree

Solution

1. Step 1: Understand balance condition

   A balanced binary tree means no node has left and right subtree heights differing by more than 1.

2. Step 2: Compare options

   The height difference between left and right subtrees of every node is at most 1 correctly states this condition. Other options describe different tree properties.

3. Final Answer:

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

4. Quick Check:

   Balance = height difference ≤ 1 [OK]

Hint: Balance means height difference ≤ 1 at every node [OK]

Common Mistakes:

• Confusing balanced with complete tree
• Thinking balance means equal number of nodes
• Assuming all nodes must have two children

2. Which TypeScript function signature correctly represents a function to check if a binary tree is balanced?

easy

A. function isBalanced(root: number): boolean

B. function isBalanced(root: TreeNode | null): boolean

C. function isBalanced(root: TreeNode): number

D. function isBalanced(root: TreeNode | null): void

Solution

1. Step 1: Identify correct parameter and return types

   The function should accept a TreeNode or null and return a boolean indicating balance.

2. Step 2: Match options

   function isBalanced(root: TreeNode | null): boolean matches this signature. Others have wrong parameter types or return types.

3. Final Answer:

   function isBalanced(root: TreeNode | null): boolean -> Option B

4. Quick Check:

   Input: TreeNode|null, Output: boolean [OK]

Hint: Check input type is TreeNode|null and output is boolean [OK]

Common Mistakes:

• Using number as input instead of TreeNode
• Returning number instead of boolean
• Using void return type

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

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val: number) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

function checkHeight(node: TreeNode | null): number {
  if (!node) return 0;
  const leftHeight = checkHeight(node.left);
  if (leftHeight === -1) return -1;
  const rightHeight = checkHeight(node.right);
  if (rightHeight === -1) return -1;
  if (Math.abs(leftHeight - rightHeight) > 1) return -1;
  return Math.max(leftHeight, rightHeight) + 1;
}

function isBalanced(root: TreeNode | null): boolean {
  return checkHeight(root) !== -1;
}

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

console.log(isBalanced(root));

medium

A. true

B. Runtime error

C. undefined

D. false

Solution

1. Step 1: Analyze tree structure

   The tree is left-heavy: root.left.left.left exists but right subtree is shallow, causing imbalance.

2. Step 2: Trace checkHeight calls

   checkHeight returns -1 when height difference > 1, so isBalanced returns false.

3. Final Answer:

   false -> Option D

4. Quick Check:

   Left-heavy deep subtree -> not balanced -> false [OK]

Hint: Deep left subtree with shallow right subtree means unbalanced [OK]

Common Mistakes:

• Assuming tree is balanced without checking height difference
• Confusing return values of checkHeight
• Ignoring the -1 sentinel for imbalance

4. Identify the bug in this TypeScript function to check if a binary tree is balanced:

function isBalanced(root: TreeNode | null): boolean {
  if (!root) return true;
  const leftHeight = getHeight(root.left);
  const rightHeight = getHeight(root.right);
  if (Math.abs(leftHeight - rightHeight) > 1) return false;
  return isBalanced(root.left) && isBalanced(root.right);
}

function getHeight(node: TreeNode | null): number {
  if (!node) return 0;
  return Math.max(getHeight(node.left), getHeight(node.right));
}

medium

A. Math.abs is used incorrectly

B. isBalanced should return true if root is null

C. getHeight does not add 1 to height, causing incorrect height calculation

D. Recursive calls in isBalanced are missing

Solution

1. Step 1: Review getHeight function

   getHeight returns max height of children but misses adding 1 for current node.

2. Step 2: Understand impact on balance check

   Without adding 1, height is underestimated, causing wrong balance results.

3. Final Answer:

   getHeight does not add 1 to height, causing incorrect height calculation -> Option C

4. Quick Check:

   Height must include current node (+1) [OK]

Hint: Always add 1 in height calculation for current node [OK]

Common Mistakes:

• Forgetting to add 1 in height function
• Misusing Math.abs
• Missing base case for null root

5. Given a binary tree, how can you efficiently check if it is balanced without recalculating heights multiple times?

hard

A. Use a bottom-up approach returning height or -1 for imbalance in one traversal

B. Calculate height of left and right subtrees separately for each node

C. Use breadth-first traversal to check balance at each level

D. Store all node heights in a global array and compare after traversal

Solution

1. Step 1: Understand inefficiency of top-down approach

   Calculating height separately for each node causes repeated work and O(n²) time.

2. Step 2: Use bottom-up approach

   Return height if balanced, else -1 for imbalance, stopping early and using one traversal.

3. Final Answer:

   Use a bottom-up approach returning height or -1 for imbalance in one traversal -> Option A

4. Quick Check:

   Bottom-up single pass = efficient balance check [OK]

Hint: Return height or -1 in one pass to avoid repeated height calls [OK]

Common Mistakes:

• Recalculating heights multiple times causing slow code
• Using breadth-first traversal which doesn't help balance check
• Storing heights globally without efficient update