Diameter of Binary Tree in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Diameter Calculation for Simple Tree

What is the output of the following C code that calculates the diameter of a binary tree?

DSA C

typedef struct Node {
    int val;
    struct Node* left;
    struct Node* right;
} Node;

int max(int a, int b) {
    return (a > b) ? a : b;
}

int height(Node* root) {
    if (!root) return 0;
    return 1 + max(height(root->left), height(root->right));
}

int diameter(Node* root) {
    if (!root) return 0;
    int leftHeight = height(root->left);
    int rightHeight = height(root->right);
    int leftDiameter = diameter(root->left);
    int rightDiameter = diameter(root->right);
    int maxDiameter = max(leftDiameter, rightDiameter);
    return max(leftHeight + rightHeight + 1, maxDiameter);
}

int main() {
    Node n4 = {4, NULL, NULL};
    Node n5 = {5, NULL, NULL};
    Node n2 = {2, &n4, &n5};
    Node n3 = {3, NULL, NULL};
    Node n1 = {1, &n2, &n3};
    int d = diameter(&n1);
    printf("%d\n", d);
    return 0;
}

A 4

B 5

C 3

D 6

💡 Hint

Think about the longest path between two nodes in the tree.

The diameter is the longest path between any two nodes. Here, the longest path goes through nodes 4 -> 2 -> 1 -> 3, which has length 4.

❓ Predict Output

intermediate

Diameter Output for Unbalanced Tree

What is the output of the diameter function for this unbalanced binary tree?

DSA C

typedef struct Node {
    int val;
    struct Node* left;
    struct Node* right;
} Node;

int max(int a, int b) {
    return (a > b) ? a : b;
}

int height(Node* root) {
    if (!root) return 0;
    return 1 + max(height(root->left), height(root->right));
}

int diameter(Node* root) {
    if (!root) return 0;
    int leftHeight = height(root->left);
    int rightHeight = height(root->right);
    int leftDiameter = diameter(root->left);
    int rightDiameter = diameter(root->right);
    int maxDiameter = max(leftDiameter, rightDiameter);
    return max(leftHeight + rightHeight + 1, maxDiameter);
}

int main() {
    Node n5 = {5, NULL, NULL};
    Node n4 = {4, &n5, NULL};
    Node n3 = {3, &n4, NULL};
    Node n2 = {2, &n3, NULL};
    Node n1 = {1, &n2, NULL};
    int d = diameter(&n1);
    printf("%d\n", d);
    return 0;
}

A 6

B 4

C 3

D 5

💡 Hint

The tree is a straight line, so diameter equals the number of nodes.

The tree is a chain of 5 nodes, so the longest path is from leaf to leaf through all nodes, length 5.

🔧 Debug

advanced

Identify the Bug in Diameter Calculation

The following code aims to calculate the diameter of a binary tree but produces incorrect results. What is the main bug?

DSA C

typedef struct Node {
    int val;
    struct Node* left;
    struct Node* right;
} Node;

int max(int a, int b) {
    return (a > b) ? a : b;
}

int height(Node* root) {
    if (!root) return 0;
    return 1 + max(height(root->left), height(root->right));
}

int diameter(Node* root) {
    if (!root) return 0;
    int leftHeight = height(root->left);
    int rightHeight = height(root->right);
    int leftDiameter = diameter(root->left);
    int rightDiameter = diameter(root->right);
    int maxDiameter = max(leftDiameter, rightDiameter);
    return max(leftHeight + rightHeight, maxDiameter);
}

A The diameter function does not check for NULL root nodes.

B The diameter calculation misses adding 1 for the current node in the path length.

C The max function is incorrectly implemented and returns the smaller value.

D The height function returns 0 for leaf nodes instead of 1.

💡 Hint

Check how the diameter is computed from heights of left and right subtrees.

The diameter should count the current node as part of the path, so it must add 1 to leftHeight + rightHeight. Missing this causes undercounting.

🚀 Application

advanced

Calculate Diameter for Given Tree Structure

Given the binary tree below, what is the diameter? Tree structure: - Root node 1 - Left child 2 with left child 4 and right child 5 - Right child 3 with right child 6 - Node 6 has left child 7 Use the diameter definition as the longest path between any two nodes.

A 7

B 5

C 6

D 8

💡 Hint

Consider the longest path that may pass through the root or deeper nodes.

The longest path is from node 7 to node 5: 7 -> 6 -> 3 -> 1 -> 2 -> 5, length 6.

🧠 Conceptual

expert

Time Complexity of Diameter Calculation

What is the time complexity of the naive diameter calculation algorithm that uses separate height calculations for each node?

A O(n^2) because height is calculated repeatedly for each node

B O(n) because each node is visited once

C O(n log n) because height calculation is logarithmic

D O(log n) because the tree height is logarithmic

💡 Hint

Consider how many times height is called for each node.

Height is called for each node and itself traverses subtrees, causing repeated work leading to O(n^2) complexity.

Practice

1. What does the diameter of a binary tree represent?

easy

A. The total number of nodes in the tree

B. The height of the tree

C. The length of the longest path between any two nodes in the tree

D. The number of leaf nodes in the tree

Solution

1. Step 1: Understand the diameter concept

   The diameter is the longest path between any two nodes, which may or may not pass through the root.

2. Step 2: Differentiate from height and node count

   Height is longest path from root to leaf, leaf count is number of leaves, total nodes is size. Diameter is distinct.

3. Final Answer:

   The length of the longest path between any two nodes in the tree -> Option C

4. Quick Check:

   Diameter = longest path length [OK]

Hint: Diameter is longest path, not height or node count [OK]

Common Mistakes:

• Confusing diameter with height
• Thinking diameter counts nodes instead of edges
• Assuming diameter always passes through root

2. Which of the following C function signatures is correct for a helper function that returns the height of a binary tree node for diameter calculation?

easy

A. int height(struct Node* root);

B. void height(struct Node root);

C. int height(Node root);

D. int height(struct Node root);

Solution

1. Step 1: Check parameter type for tree node

   In C, tree nodes are passed as pointers to struct Node, so parameter should be struct Node* root.

2. Step 2: Check return type and parameter correctness

   Height returns int, so return type int is correct. Passing pointer is correct, not struct by value.

3. Final Answer:

   int height(struct Node* root); -> Option A

4. Quick Check:

   Pointer parameter and int return type [OK]

Hint: Use pointer to struct Node for tree nodes in C [OK]

Common Mistakes:

• Passing struct Node by value instead of pointer
• Using void return type instead of int
• Missing struct keyword in parameter

3. What is the output of the following C code snippet that calculates the diameter of a binary tree?

struct Node {
  int data;
  struct Node* left;
  struct Node* right;
};

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

int diameter(struct Node* root) {
  if (!root) return 0;
  int leftHeight = height(root->left);
  int rightHeight = height(root->right);
  int leftDiameter = diameter(root->left);
  int rightDiameter = diameter(root->right);
  int maxDiameter = leftHeight + rightHeight + 1;
  if (leftDiameter > maxDiameter) maxDiameter = leftDiameter;
  if (rightDiameter > maxDiameter) maxDiameter = rightDiameter;
  return maxDiameter;
}

int main() {
  struct Node n1 = {1, NULL, NULL};
  struct Node n2 = {2, NULL, NULL};
  struct Node n3 = {3, &n1, &n2};
  printf("%d", diameter(&n3));
  return 0;
}

medium

A. 1

B. 2

C. 4

D. 3

Solution

1. Step 1: Analyze tree structure

   Root n3 has two children n1 and n2, both leaves. Height of left and right subtree is 1 each.

2. Step 2: Calculate diameter

   Diameter = max of (leftHeight + rightHeight + 1, leftDiameter, rightDiameter) = max(1+1+1, 1, 1) = 3.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Diameter = 3 nodes in longest path [OK]

Hint: Diameter = max(leftHeight + rightHeight + 1, leftDiameter, rightDiameter) [OK]

Common Mistakes:

• Confusing diameter with height
• Forgetting to add 1 for root node in diameter
• Using height instead of diameter in comparisons

4. Identify the bug in this C function that attempts to compute the diameter of a binary tree:

int diameter(struct Node* root) {
  if (!root) return 0;
  int leftHeight = height(root->left);
  int rightHeight = height(root->right);
  int leftDiameter = diameter(root->left);
  int rightDiameter = diameter(root->right);
  int maxDiameter = leftHeight + rightHeight;
  if (leftDiameter > maxDiameter) maxDiameter = leftDiameter;
  if (rightDiameter > maxDiameter) maxDiameter = rightDiameter;
  return maxDiameter;
}

medium

A. height function is not called correctly

B. maxDiameter should be leftHeight + rightHeight + 1

C. Function should return height, not diameter

D. Missing base case for leaf nodes

Solution

1. Step 1: Check diameter calculation formula

   Diameter includes edges between nodes, so it should add 1 for the root node connecting left and right paths.

2. Step 2: Identify missing +1 in maxDiameter

   Current code misses +1, so maxDiameter undercounts by 1.

3. Final Answer:

   maxDiameter should be leftHeight + rightHeight + 1 -> Option B

4. Quick Check:

   Diameter = leftHeight + rightHeight + 1 [OK]

Hint: Add 1 to sum of heights for diameter at root [OK]

Common Mistakes:

• Omitting +1 in diameter calculation
• Confusing height and diameter returns
• Not handling null root base case

5. Given a binary tree where each node stores an integer, how would you modify the diameter calculation to find the diameter based on the sum of node values along the longest path instead of the number of nodes?

hard

A. Change height function to return max sum of node values from root to leaf, and update diameter to sum left and right max sums plus root's value

B. Use the existing diameter function but multiply the result by the root node's value

C. Replace all node values with 1 and then calculate diameter normally

D. Calculate diameter normally and then add all node values in the tree separately

Solution

1. Step 1: Modify height to return max sum path

   Instead of counting nodes, height returns max sum of node values from root to leaf recursively.

2. Step 2: Update diameter to sum left and right max sums plus root value

   Diameter becomes max of leftDiameter, rightDiameter, and leftSum + rightSum + root's value.

3. Final Answer:

   Change height function to return max sum of node values from root to leaf, and update diameter to sum left and right max sums plus root's value -> Option A

4. Quick Check:

   Diameter by sum = max path sum through root [OK]

Hint: Use max sum path instead of node count for diameter sum [OK]

Common Mistakes:

• Multiplying diameter by root value incorrectly
• Ignoring node values in diameter calculation
• Adding all node values separately instead of path sums