Check if Binary Tree is Balanced in DSA Javascript - Time & Space Complexity

Time Complexity: Check if Binary Tree is Balanced

O(n)

Understanding Time Complexity

We want to know how long it takes to check if a binary tree is balanced.

Specifically, how the time grows as the tree gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code checks if a binary tree is balanced by measuring heights and stopping early if imbalance is found.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Each node is visited once to calculate height and check balance.

How Execution Grows With Input

As the tree grows, the function visits each node once, so operations grow directly with number of nodes.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: Operations increase linearly as tree size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to check balance grows in direct proportion to the number of nodes in the tree.

Common Mistake

[X] Wrong: "The function only checks the root, so it runs in constant time O(1)."

[OK] Correct: The function must visit every node to check heights and balance, so it depends on tree size.

Interview Connect

Understanding this helps you explain how recursion explores trees efficiently and why early stopping matters.

Self-Check

"What if the function did not stop early on imbalance and always checked all nodes? How would the time complexity change?"

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