Check if Binary Tree is Balanced in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does it mean for a binary tree to be balanced?

A binary tree is balanced if for every node, the height difference between its left and right subtrees is at most 1.

beginner

Why is checking balance important in binary trees?

Balanced trees ensure operations like search, insert, and delete run efficiently, typically in O(log n) time.

intermediate

What is the main idea behind the efficient algorithm to check if a binary tree is balanced?

Use a bottom-up approach that returns the height of subtrees and checks balance at the same time to avoid repeated height calculations.

intermediate

What does the function return if the subtree is not balanced in the efficient approach?

It returns -1 to indicate the subtree is not balanced, which helps stop further unnecessary checks.

beginner

In the context of checking if a binary tree is balanced, what is the height of a null node?

The height of a null node is defined as 0, meaning it contributes no height to the tree.

What is the maximum allowed height difference between left and right subtrees for a node to be balanced?

A 0

B 2

C 1

D 3

A balanced binary tree requires the height difference to be at most 1.

Which approach is more efficient to check if a binary tree is balanced?

A Level order traversal only

B Top-down approach calculating height repeatedly

C Inorder traversal only

D Bottom-up approach returning height and balance status

The bottom-up approach avoids repeated height calculations and stops early if imbalance is found.

What does a height function return when it detects an unbalanced subtree?

A 0

B -1

C 1

D Height of the subtree

Returning -1 signals that the subtree is unbalanced.

What is the height of an empty (null) subtree?

A 0

B -1

C 1

D Undefined

An empty subtree has height 0.

If a binary tree is balanced, what is the time complexity of checking it using the bottom-up approach?

A O(n)

B O(n^2)

C O(n log n)

D O(log n)

The bottom-up approach visits each node once, so it runs in O(n) time.

Explain how the bottom-up approach checks if a binary tree is balanced.

Think about how you can combine height calculation and balance checking in one function.

Describe why the top-down approach to check balance is less efficient than the bottom-up approach.

Consider how many times each node is processed in both approaches.

Practice

1. What does it mean for a binary tree to be balanced?

easy

A. All nodes have two children

B. The height difference between left and right subtrees of every node is at most 1

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 the height difference between left and right subtrees of every node is at most 1.

2. Step 2: Compare other options

   Options B, C, 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 B

4. Quick Check:

   Balance means 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 of the following C++ function signatures is correct to check if a binary tree is balanced?

easy

A. bool isBalanced(TreeNode* root);

B. int isBalanced(TreeNode root);

C. void isBalanced(TreeNode* root);

D. bool isBalanced(TreeNode root);

Solution

1. Step 1: Identify correct parameter type

   The function should accept a pointer to the root node (TreeNode* root) to access the tree.

2. Step 2: Identify correct return type

   It should return bool to indicate if the tree is balanced or not.

3. Final Answer:

   bool isBalanced(TreeNode* root); -> Option A

4. Quick Check:

   Pointer parameter + bool return = correct signature [OK]

Hint: Use pointer parameter and bool return type [OK]

Common Mistakes:

• Passing node by value instead of pointer
• Using int or void as return type
• Missing pointer symbol * in parameter

3. What will be the output of the following code snippet?

struct TreeNode {
  int val;
  TreeNode* left;
  TreeNode* right;
  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

int height(TreeNode* root) {
  if (!root) return 0;
  int leftHeight = height(root->left);
  int rightHeight = height(root->right);
  return 1 + (leftHeight > rightHeight ? leftHeight : rightHeight);
}

bool isBalanced(TreeNode* root) {
  if (!root) return true;
  int leftHeight = height(root->left);
  int rightHeight = height(root->right);
  if (abs(leftHeight - rightHeight) > 1) return false;
  return isBalanced(root->left) && isBalanced(root->right);
}

int main() {
  TreeNode* 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);
  std::cout << (isBalanced(root) ? "Balanced" : "Not Balanced") << std::endl;
  return 0;
}

medium

A. Not Balanced

B. Balanced

C. Compilation Error

D. Runtime Error

Solution

1. Step 1: Calculate heights of subtrees

   Left subtree has height 3 (nodes 2->4->5), right subtree height is 1 (node 3).

2. Step 2: Check balance condition

   Height difference at root is 3 - 1 = 2, which is greater than 1, so tree is not balanced.

3. Final Answer:

   Not Balanced -> Option A

4. Quick Check:

   Height difference > 1 means Not Balanced [OK]

Hint: Check height difference > 1 means unbalanced [OK]

Common Mistakes:

• Ignoring height difference at root
• Assuming tree is balanced if only some nodes are balanced
• Confusing height with number of nodes

4. Identify the bug in this code snippet that checks if a binary tree is balanced:

int height(TreeNode* root) {
  if (!root) return 0;
  int leftHeight = height(root->left);
  int rightHeight = height(root->right);
  return 1 + (leftHeight > rightHeight ? leftHeight : rightHeight);
}

bool isBalanced(TreeNode* root) {
  if (!root) return true;
  int leftHeight = height(root->left);
  int rightHeight = height(root->right);
  if (abs(leftHeight - rightHeight) > 1) return false;
  isBalanced(root->left);
  isBalanced(root->right);
  return true;
}

medium

A. Base case in height is incorrect

B. height function returns wrong height

C. abs function is not included

D. Missing return statements for recursive calls in isBalanced

Solution

1. Step 1: Analyze isBalanced recursive calls

   Recursive calls to isBalanced(root->left) and isBalanced(root->right) are made but their results are ignored.

2. Step 2: Fix missing return statements

   Must return the logical AND of these calls to correctly check balance in subtrees.

3. Final Answer:

   Missing return statements for recursive calls in isBalanced -> Option D

4. Quick Check:

   Recursive calls must return and combine results [OK]

Hint: Return recursive calls combined with && [OK]

Common Mistakes:

• Ignoring return values of recursive calls
• Returning true too early
• Incorrect base case handling

5. Given a binary tree, how can you optimize the balance check to avoid repeated height calculations?

hard

A. Use breadth-first traversal to check balance

B. Calculate height separately for left and right subtrees multiple times

C. Use a helper function that returns both height and balance status in one traversal

D. Check balance only at the root node

Solution

1. Step 1: Understand repeated height calculation problem

   Calculating height separately for each node causes repeated work and increases time complexity.

2. Step 2: Use combined helper function

   A helper function can return both height and balance status together, allowing one traversal to check balance efficiently.

3. Final Answer:

   Use a helper function that returns both height and balance status in one traversal -> Option C

4. Quick Check:

   Combine height and balance check in one pass [OK]

Hint: Combine height and balance check in one function [OK]

Common Mistakes:

• Calculating height multiple times causing O(n²) time
• Checking balance only at root misses subtree imbalance
• Using BFS does not directly help balance check