Check if Binary Tree is Balanced in DSA Typescript - 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: TreeNode | null): boolean {
  function height(node: TreeNode | null): number {
    if (!node) return 0;
    const leftHeight = height(node.left);
    if (leftHeight === -1) return -1;
    const rightHeight = height(node.right);
    if (rightHeight === -1) return -1;
    if (Math.abs(leftHeight - rightHeight) > 1) return -1;
    return Math.max(leftHeight, rightHeight) + 1;
  }
  return height(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 to visit 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.

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

Pattern observation: The work grows directly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to check balance grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The function checks balance in O(n log n) time because it calculates height separately for each node."

[OK] Correct: This code calculates height once per node using a single recursion, avoiding repeated height calculations, so it runs in O(n) time.

Interview Connect

Understanding this time complexity helps you explain efficient tree checks clearly and confidently in interviews.

Self-Check

"What if the code calculated height separately for left and right subtrees without early stopping? How would the time complexity change?"

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