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

Time Complexity: Diameter of Binary Tree

O(n)

Understanding Time Complexity

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

Specifically, how does the work increase when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public class TreeNode {
    public int val;
    public TreeNode left, right;
    public TreeNode(int val=0) { this.val = val; }
}

int diameter = 0;

int Depth(TreeNode node) {
    if (node == null) return 0;
    int leftDepth = Depth(node.left);
    int rightDepth = Depth(node.right);
    diameter = Math.Max(diameter, leftDepth + rightDepth);
    return Math.Max(leftDepth, rightDepth) + 1;
}

// Call Depth(root) to compute diameter

This code finds the longest path between any two nodes in a binary tree by computing depths recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls to Depth for each node.
• How many times: Each node is visited once to calculate depth and update diameter.

How Execution Grows With Input

As the number of nodes increases, the function visits each node once.

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

Pattern observation: The work grows directly with the number of nodes, so doubling nodes roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the diameter grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The diameter calculation takes more than linear time because it checks many paths repeatedly."

[OK] Correct: The code visits each node once, calculating depth and updating diameter on the fly, so it does not repeat work for the same nodes.

Interview Connect

Understanding this linear time solution shows you can efficiently handle tree problems using recursion and careful tracking, a key skill in many coding challenges.

Self-Check

"What if we changed the code to compute diameter by separately computing depths for each node multiple times? How would the time complexity change?"

Practice

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

easy

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

B. The height of the tree from root to the deepest leaf

C. The total number of 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 the longest path from root to leaf, node count is total nodes, and leaf count is number of leaves, none of which define diameter.

3. Final Answer:

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

4. Quick Check:

   Diameter = longest path between nodes [OK]

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

Common Mistakes:

• Confusing diameter with tree height
• Thinking diameter counts nodes instead of edges
• Assuming diameter must pass through root

2. Which of the following C# method signatures is correct for a function that calculates the diameter of a binary tree node?

easy

A. Diameter public int(TreeNode root)

B. int Diameter(TreeNode root) void

C. public void Diameter(TreeNode root) int

D. public int Diameter(TreeNode root)

Solution

1. Step 1: Review C# method signature syntax

   In C#, method signature starts with access modifier, return type, method name, and parameters in parentheses.

2. Step 2: Check each option

   public int Diameter(TreeNode root) follows correct syntax: public int Diameter(TreeNode root). Others have incorrect order or missing parts.

3. Final Answer:

   public int Diameter(TreeNode root) -> Option D

4. Quick Check:

   Correct C# method syntax = public int Diameter(TreeNode root) [OK]

Hint: Method signature: access modifier + return type + name + parameters [OK]

Common Mistakes:

• Placing return type after parameters
• Omitting access modifier or return type
• Incorrect order of keywords

3. Given the following binary tree code snippet, what is the diameter of the tree?

class TreeNode {
  public int val;
  public TreeNode left, right;
  public TreeNode(int v) { val = v; }
}

// Tree structure:
//       1
//      / \
//     2   3
//    / 
//   4

TreeNode root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(4);

int diameter = CalculateDiameter(root); // Assume CalculateDiameter correctly implemented
Console.WriteLine(diameter);

medium

A. 3

B. 2

C. 4

D. 5

Solution

1. Step 1: Visualize the tree and longest path

   The longest path is from node 4 to node 3: path length = edges 4->2->1->3 = 3 edges.

2. Step 2: Calculate diameter as number of edges

   Diameter counts edges, so longest path length is 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Longest path edges = 3 [OK]

Hint: Count edges on longest path, not nodes [OK]

Common Mistakes:

• Counting nodes instead of edges
• Assuming diameter passes through root only
• Ignoring one side of the tree

4. Identify the bug in this C# code snippet for calculating the diameter of a binary tree:

int maxDiameter = 0;
int Depth(TreeNode node) {
  if (node == null) return 0;
  int left = Depth(node.left);
  int right = Depth(node.right);
  maxDiameter = Math.Max(maxDiameter, left + right + 1);
  return Math.Max(left, right) + 1;
}

int Diameter(TreeNode root) {
  Depth(root);
  return maxDiameter;
}

medium

A. The base case should return 1 instead of 0

B. Diameter should be maxDiameter - 1 because diameter counts edges, not nodes

C. maxDiameter should be reset inside Depth method

D. Depth method should return left + right instead of max

Solution

1. Step 1: Understand diameter counting edges vs nodes

   The code adds left + right + 1 (nodes), but diameter counts edges, so subtract 1.

2. Step 2: Fix return value in Diameter method

   Return maxDiameter - 1 to convert node count to edge count for diameter.

3. Final Answer:

   Diameter should be maxDiameter - 1 because diameter counts edges, not nodes -> Option B

4. Quick Check:

   Diameter edges = nodes - 1 [OK]

Hint: Diameter = nodes on path - 1 (edges) [OK]

Common Mistakes:

• Returning maxDiameter directly without subtracting 1
• Changing base case return value incorrectly
• Resetting maxDiameter inside recursive calls

5. You want to find the diameter of a binary tree but also return the path nodes forming that diameter. Which approach correctly extends the diameter calculation?

hard

A. Calculate diameter first, then do a separate BFS to find the path

B. Store all paths from root to leaves and pick the longest one

C. Modify the recursive function to return both depth and path, updating max path when diameter increases

D. Use a queue to traverse level by level and count nodes

Solution

1. Step 1: Understand diameter and path relationship

   Diameter is longest path between any two nodes, so tracking path during recursion helps capture it.

2. Step 2: Extend recursion to return depth and path

   Modify recursive calls to return depth and path list; update max path when a longer diameter is found.

3. Step 3: Compare other options

   Calculate diameter first, then do a separate BFS to find the path is inefficient; C only finds longest root-to-leaf path, not diameter; D is unrelated to diameter.

4. Final Answer:

   Modify the recursive function to return both depth and path, updating max path when diameter increases -> Option C

5. Quick Check:

   Track depth and path together for diameter path [OK]

Hint: Return depth and path in recursion to track diameter path [OK]

Common Mistakes:

• Trying to find path after diameter without tracking during recursion
• Confusing longest root-to-leaf path with diameter
• Using BFS level order traversal which doesn't find diameter