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.