Diameter of Binary Tree in DSA C++ - Time & Space Complexity

Time Complexity: Diameter of Binary Tree

O(n²)

Understanding Time Complexity

We want to understand how the time needed to find the diameter of a binary tree changes as the tree grows.

The question is: how does the work increase when the tree has more nodes?

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 + max(leftHeight, rightHeight);
}

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 max(leftHeight + rightHeight, max(leftDiameter, rightDiameter));
}
    

This code finds the diameter by computing heights and diameters of subtrees recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls to height and diameter functions.
• How many times: For each node, height is called multiple times leading to repeated work.

How Execution Grows With Input

As the number of nodes (n) increases, the height function is called many times for the same nodes, causing repeated work.

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

Pattern observation: The work grows roughly with the square of the number of nodes because height is recomputed many times.

Final Time Complexity

Time Complexity: O(n²)

This means the time needed grows roughly with the square of the number of nodes in the tree.

Common Mistake

[X] Wrong: "The diameter function runs in linear time because it visits each node once."

[OK] Correct: The height function is called repeatedly for the same nodes, causing extra work that adds up to quadratic time.

Interview Connect

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

Self-Check

"What if we compute height once for each node and store it? How would the time complexity change?"

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