Why Check if Binary Tree is Balanced in DSA C++?

The Big Idea

Discover how a clever trick saves you from endless counting and guessing in trees!

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 each branch's height and comparing takes a lot of time and is easy to make mistakes. You might forget to check some parts or miscount the levels, leading to wrong conclusions.

The Solution

Using a smart method, you can check the balance of the tree by looking at each node once, calculating heights, and deciding if the tree is balanced quickly and without errors.

Before vs After

✗ Before

int height(Node* node) {
  if (!node) return 0;
  return 1 + max(height(node->left), height(node->right));
}
bool isBalanced(Node* node) {
  if (!node) return true;
  int leftHeight = height(node->left);
  int rightHeight = height(node->right);
  if (abs(leftHeight - rightHeight) > 1) return false;
  return isBalanced(node->left) && isBalanced(node->right);
}

✓ After

int checkHeight(Node* node) {
  if (!node) return 0;
  int leftHeight = checkHeight(node->left);
  if (leftHeight == -1) return -1;
  int rightHeight = checkHeight(node->right);
  if (rightHeight == -1) return -1;
  if (abs(leftHeight - rightHeight) > 1) return -1;
  return 1 + max(leftHeight, rightHeight);
}
bool isBalanced(Node* root) {
  return checkHeight(root) != -1;
}

What It Enables

This method lets you quickly and correctly find out if a tree is balanced, helping keep data organized and efficient.

Real Life Example

In computer games, balanced trees help manage objects so the game runs smoothly without delays or glitches.

Key Takeaways

Manual checking is slow and error-prone.

Smart recursive method checks balance in one pass.

Balanced trees improve performance and organization.

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