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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of diameter calculation for a simple binary tree

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

DSA C++

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

int maxDiameter = 0;

int depth(TreeNode* node) {
    if (!node) return 0;
    int leftDepth = depth(node->left);
    int rightDepth = depth(node->right);
    maxDiameter = std::max(maxDiameter, leftDepth + rightDepth);
    return 1 + std::max(leftDepth, rightDepth);
}

int diameterOfBinaryTree(TreeNode* root) {
    maxDiameter = 0;
    depth(root);
    return maxDiameter;
}

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->right = new TreeNode(5);
    int result = diameterOfBinaryTree(root);
    std::cout << result << std::endl;
    return 0;
}

A 4

B 2

C 3

D 5

💡 Hint

The diameter is the longest path between any two nodes, which may pass through the root.

The longest path goes through nodes 4 -> 2 -> 1 -> 3 or 5 -> 2 -> 1 -> 3, which has length 3 edges.

❓ Predict Output

intermediate

Diameter output for a skewed binary tree

What will be the output of the diameterOfBinaryTree function for the following skewed binary tree?

DSA C++

TreeNode* root = new TreeNode(1);
root->right = new TreeNode(2);
root->right->right = new TreeNode(3);
root->right->right->right = new TreeNode(4);
int result = diameterOfBinaryTree(root);
std::cout << result << std::endl;

A 0

B 4

C 1

D 3

💡 Hint

In a skewed tree, the diameter is the length of the longest path which is the number of edges in the chain.

The tree is a straight line with 4 nodes, so the diameter is 3 edges.

🔧 Debug

advanced

Identify the error in diameter calculation code

What error will the following code produce when calculating the diameter of a binary tree?

DSA C++

int maxDiameter = 0;

int depth(TreeNode* node) {
    if (node == nullptr) return 0;
    int leftDepth = depth(node->left);
    int rightDepth = depth(node->right);
    maxDiameter = std::max(maxDiameter, leftDepth * rightDepth);
    return 1 + std::max(leftDepth, rightDepth);
}

int diameterOfBinaryTree(TreeNode* root) {
    maxDiameter = 0;
    depth(root);
    return maxDiameter;
}

A The diameter will be incorrect because it uses multiplication instead of addition

B Compilation error due to missing return statement

C Runtime error due to null pointer dereference

D No error, code works correctly

💡 Hint

Check how diameter is calculated from left and right depths.

Diameter is the sum of left and right depths, not their product. Using multiplication gives wrong results.

🧠 Conceptual

advanced

Understanding diameter definition in binary trees

Which of the following best describes the diameter of a binary tree?

A The number of edges on the longest path between any two nodes in the tree

B The height of the tree plus one

C The maximum depth of the left subtree

D The number of nodes on the longest path between any two nodes in the tree

💡 Hint

Diameter counts edges, not nodes.

Diameter is defined as the number of edges on the longest path between any two nodes.

🚀 Application

expert

Calculate diameter after modifying the tree

Given the initial tree below, what is the diameter after adding a new node as the left child of node 5? Initial tree: 1 ├─ 2 │ ├─ 4 │ └─ 5 └─ 3 Add node 6 as left child of node 5. What is the new diameter?

DSA C++

TreeNode* root = new TreeNode(1);
root->left = new TreeNode(2);
root->right = new TreeNode(3);
root->left->left = new TreeNode(4);
root->left->right = new TreeNode(5);
root->left->right->left = new TreeNode(6);
int result = diameterOfBinaryTree(root);
std::cout << result << std::endl;

A 5

B 4

C 3

D 6

💡 Hint

Consider the longest path after adding the new node 6.

The longest path is now 6 -> 5 -> 2 -> 1 -> 3, which has 4 edges.

Practice

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

easy

A. The longest path between any two nodes in the tree

B. The height of the tree

C. The number of leaf nodes

D. The total number of nodes in the tree

Solution

1. Step 1: Understand the diameter definition

   The diameter is the longest path between any two nodes, not just from root to leaf.

2. Step 2: Differentiate from height and node count

   Height is longest root-to-leaf path; node count and leaf count are unrelated to diameter.

3. Final Answer:

   The longest path between any two nodes in the tree -> Option A

4. Quick Check:

   Diameter = longest node-to-node path [OK]

Hint: Diameter is longest path between any two nodes [OK]

Common Mistakes:

• Confusing diameter with tree height
• Thinking diameter counts nodes instead of edges
• Assuming diameter is number of leaf nodes

2. Which of the following is the correct C++ function signature to compute the diameter of a binary tree node?

easy

A. int diameter(TreeNode& root);

B. void diameter(TreeNode root);

C. int diameter(TreeNode root);

D. int diameter(TreeNode* root);

Solution

1. Step 1: Understand pointer usage in tree nodes

   Binary trees are usually handled with pointers to nodes, so TreeNode* is correct.

2. Step 2: Check return type and parameter correctness

   Diameter returns an int, and parameter should be pointer, not value or reference for typical tree traversal.

3. Final Answer:

   int diameter(TreeNode* root); -> Option D

4. Quick Check:

   Pointer parameter with int return [OK]

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

Common Mistakes:

• Passing node by value instead of pointer
• Using void return type instead of int
• Passing node by reference unnecessarily

3. What is the output of the following code snippet computing diameter?

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;
  return 1 + std::max(height(root->left), height(root->right));
}

int diameter(TreeNode* 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);
  return std::max(leftHeight + rightHeight, std::max(leftDiameter, rightDiameter));
}

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->right = new TreeNode(5);
  std::cout << diameter(root);
  return 0;
}

medium

A. 6

B. 4

C. 3

D. 5

Solution

1. Step 1: Calculate heights of left and right subtrees

   Left subtree height: max height of nodes 2->4 or 2->5 is 2; right subtree height (node 3) is 1.

2. Step 2: Calculate diameter candidates and choose max

   Diameter candidates: leftHeight + rightHeight = 2 + 1 = 3; leftDiameter and rightDiameter are computed recursively. LeftDiameter is 2 (path 4-2-5), rightDiameter is 0. Max is 3.

3. Final Answer:

   3 -> Option C

4. Quick Check:

   Longest path length = 3 edges [OK]

Hint: Sum left and right subtree heights for diameter [OK]

Common Mistakes:

• Confusing height with diameter
• Counting nodes instead of edges
• Ignoring recursive diameter calls

4. Identify the bug in this diameter function and select the fix:

int diameter(TreeNode* 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);
  return leftHeight + rightHeight;
}

medium

A. Return max of leftDiameter and rightDiameter only

B. Return max of leftHeight + rightHeight and leftDiameter, rightDiameter

C. Return sum of leftDiameter and rightDiameter

D. Return height of root node

Solution

1. Step 1: Analyze current return statement

   Currently returns only leftHeight + rightHeight, ignoring recursive diameters from subtrees.

2. Step 2: Correct return to consider all diameter candidates

   Diameter is max of leftHeight + rightHeight, leftDiameter, and rightDiameter.

3. Final Answer:

   Return max of leftHeight + rightHeight and leftDiameter, rightDiameter -> Option B

4. Quick Check:

   Include all diameter candidates [OK]

Hint: Return max of height sum and subtree diameters [OK]

Common Mistakes:

• Ignoring recursive diameter calls
• Returning sum of diameters incorrectly
• Returning height instead of diameter

5. Given a binary tree where each node has a value, how would you modify the diameter function to return the maximum sum of node values on any path between two nodes instead of the number of edges?

hard

A. Change height function to return max sum of node values and update diameter to use sums instead of heights

B. Use the same diameter function but multiply the result by node values

C. Calculate diameter normally and add all node values in the tree

D. Return the maximum node value in the tree as diameter

Solution

1. Step 1: Modify height to return max sum of node values

   Instead of counting edges, height returns max sum from node to leaf including node values.

2. Step 2: Update diameter to use sums from left and right subtrees

   Diameter becomes max of left sum + right sum and recursive diameters, reflecting max path sum.

3. Final Answer:

   Change height function to return max sum of node values and update diameter to use sums instead of heights -> Option A

4. Quick Check:

   Use sums in height and diameter for max path value [OK]

Hint: Use sums in height and diameter for max path value [OK]

Common Mistakes:

• Multiplying diameter by node values incorrectly
• Adding all node values regardless of path
• Returning max node value instead of path sum