Check if Binary Tree is Balanced in DSA Javascript - Build from Scratch

Check if Binary Tree is Balanced

📖 Scenario: You are working on a program that manages a family tree. To keep the tree easy to search and update, you want to check if the tree is balanced. A balanced tree means the left and right branches of every node are about the same height, so it doesn't lean too much to one side.

🎯 Goal: Build a program that creates a simple binary tree, sets a limit for height difference, checks if the tree is balanced, and prints the result.

📋 What You'll Learn

Create a binary tree using nodes with value, left, and right properties

Create a variable maxAllowedDifference to set the maximum height difference allowed between left and right subtrees

Write a function isBalanced that checks if the binary tree is balanced based on maxAllowedDifference

Print true or false depending on whether the tree is balanced

💡 Why This Matters

🌍 Real World

Balanced trees are important in databases and search engines to keep data easy to find quickly.

💼 Career

Understanding tree balance helps in roles like software development, data engineering, and system design where efficient data structures matter.

1

Create the Binary Tree

Create a binary tree with a root node called root. The root should have value 10, a left child with value 5, and a right child with value 15. Each node should be an object with value, left, and right properties. The left and right children of the leaf nodes should be null.

DSA Javascript

// Create the binary tree with root, left, and right nodes
// Your code here

Hint

Think of each node as a small box with three parts: value, left, and right. Leaf nodes have left and right as null.

2

Set Maximum Allowed Height Difference

Create a variable called maxAllowedDifference and set it to 1. This will be the maximum allowed difference in height between the left and right subtrees for the tree to be considered balanced.

DSA Javascript

const root = {
  value: 10,
  left: { value: 5, left: null, right: null },
  right: { value: 15, left: null, right: null }
};

// Create maxAllowedDifference variable and set to 1
// Your code here

Hint

This variable helps decide how much difference in height is okay between left and right branches.

3

Write the Function to Check if Tree is Balanced

Write a function called isBalanced that takes a node as input and returns true if the tree rooted at that node is balanced, or false otherwise. Use maxAllowedDifference to check the height difference between left and right subtrees. You can write a helper function inside isBalanced to calculate the height of a subtree.

DSA Javascript

const root = {
  value: 10,
  left: { value: 5, left: null, right: null },
  right: { value: 15, left: null, right: null }
};

const maxAllowedDifference = 1;

// Write function isBalanced(node) below
// Your code here

Hint

Use a helper function to find height. Check difference at current node, then check left and right subtrees recursively.

4

Print if the Tree is Balanced

Write a console.log statement to print the result of calling isBalanced with root as the argument.

DSA Javascript

const root = {
  value: 10,
  left: { value: 5, left: null, right: null },
  right: { value: 15, left: null, right: null }
};

const maxAllowedDifference = 1;

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

  if (node === null) return true;

  const leftHeight = height(node.left);
  const rightHeight = height(node.right);

  if (Math.abs(leftHeight - rightHeight) > maxAllowedDifference) {
    return false;
  }

  return isBalanced(node.left) && isBalanced(node.right);
}

// Print if the tree is balanced
// Your code here

Hint

Call isBalanced(root) and print the result using console.log.

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