Why Check if Binary Tree is Balanced in DSA Javascript?

The Big Idea

Discover how a simple trick saves you from endless measuring and guessing if a tree is balanced!

The Scenario

Imagine you have a family tree drawn on paper. You want to check if the tree looks balanced, meaning no side is too tall compared to the other. Doing this by measuring each branch manually is tiring and confusing.

The Problem

Manually checking if a tree is balanced means measuring every branch length and comparing them. This is slow, easy to make mistakes, and hard to keep track of all parts, especially for big trees.

The Solution

Using a smart method, we can check balance by looking at each node once, calculating heights, and deciding if the tree is balanced without repeating work. This saves time and avoids errors.

Before vs After

✗ Before

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);
}

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

✓ After

function checkBalance(node) {
  if (!node) return 0;
  let leftHeight = checkBalance(node.left);
  if (leftHeight === -1) return -1;
  let rightHeight = checkBalance(node.right);
  if (rightHeight === -1) return -1;
  if (Math.abs(leftHeight - rightHeight) > 1) return -1;
  return 1 + Math.max(leftHeight, rightHeight);
}

function isBalanced(root) {
  return checkBalance(root) !== -1;
}

What It Enables

This method lets us quickly and reliably know if a tree is balanced, which helps keep data organized and fast to search.

Real Life Example

In computer games, balanced trees help manage objects so the game runs smoothly without delays caused by unbalanced data.

Key Takeaways

Manual checking is slow and error-prone.

Smart recursive method checks balance efficiently.

Balanced trees improve performance in many applications.

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