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

Time Complexity: Check if Binary Tree is Balanced

O(n²)

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.

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

This code checks if a binary tree is balanced by comparing heights of subtrees recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The height function is called repeatedly for each node.
• How many times: For each node, height is called again on its children, causing repeated calculations.

How Execution Grows With Input

As the tree size grows, the number of height calculations grows quickly because each node's height is recalculated many times.

Input Size (n)	Approx. Operations
10	About 100 calls
100	About 10,000 calls
1000	About 1,000,000 calls

Pattern observation: The operations grow roughly with the square of the number of nodes.

Final Time Complexity

Time Complexity: O(n²)

This means the time to check balance grows very fast as the tree gets bigger, roughly like the number of nodes squared.

Common Mistake

[X] Wrong: "The height function is called only once per node, so the time is O(n)."

[OK] Correct: Actually, height is called many times for the same nodes during recursion, causing repeated work and increasing time.

Interview Connect

Understanding this helps you explain why naive recursion can be slow and how to improve it, a key skill in interviews.

Self-Check

"What if we modified the code to calculate height and balance in the same recursion? How would the time complexity change?"

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