Check if Binary Tree is Balanced in DSA C++ - Build from Scratch

Check if Binary Tree is Balanced

📖 Scenario: Imagine you are building a program to check if a family tree is well balanced. A balanced tree means no branch is too long compared to others, making it easier to find family members quickly.

🎯 Goal: You will create a binary tree, add a helper variable, write a function to check if the tree is balanced, and finally print the result.

📋 What You'll Learn

Create a binary tree with the exact nodes and structure given

Add a helper function to calculate height

Write a function called isBalanced that returns true if the tree is balanced

Print "Balanced" if the tree is balanced, otherwise print "Not Balanced"

💡 Why This Matters

🌍 Real World

Balanced trees are used in databases and file systems to keep data organized for fast searching and updating.

💼 Career

Understanding tree balance helps in roles like software development, data engineering, and system design where efficient data structures are crucial.

1

Create the Binary Tree Structure

Create a struct called Node with an int data, and two pointers left and right. Then create the exact binary tree below by creating Node pointers: root with value 1, root's left child 2, root's right child 3, left child of 2 is 4, right child of 2 is 5.

DSA C++

// Define struct Node
// Create nodes and link them to form the tree
// Your code here

Hint

Define the Node struct first. Then create nodes with new Node(value) and link them using ->left and ->right.

2

Add a Helper Function to Calculate Height

Add a function called height that takes a Node* called root and returns an int. It should return 0 if root is nullptr. Otherwise, return 1 plus the maximum height of the left and right subtrees.

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
root->left->right = new Node(5);

// Add height function below
// Your code here

Hint

Use recursion to find the height of left and right subtrees, then return 1 plus the bigger one.

3

Write the isBalanced Function

Write a function called isBalanced that takes a Node* called root and returns a bool. It should return true if the tree is balanced. A tree is balanced if for every node, the difference between the height of left and right subtrees is at most 1, and both left and right subtrees are balanced.

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
root->left->right = new Node(5);

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

// Write isBalanced function below
// Your code here

Hint

Check height difference at current node and recursively check left and right subtrees.

4

Print if the Tree is Balanced

Write a main function that calls isBalanced(root). If it returns true, print "Balanced". Otherwise, print "Not Balanced".

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
root->left->right = new Node(5);

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

bool isBalanced(Node* root) {
    if (root == nullptr) 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() {
    // Print "Balanced" or "Not Balanced" based on isBalanced(root)
    // Your code here
}

Hint

Use cout to print the result based on isBalanced(root).

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