DP on Trees Diameter of Tree in DSA C# - Build from Scratch

DP on Trees: Diameter of Tree

📖 Scenario: You are working with a tree structure representing a network of connected points. You want to find the longest path between any two points in this network. This longest path is called the diameter of the tree.Imagine the tree as a set of connected cities with roads. The diameter is the longest road trip you can take between two cities without revisiting any city.

🎯 Goal: Build a program that calculates the diameter of a tree using dynamic programming on trees. You will create the tree, set up helper variables, write the logic to find the diameter, and finally print the diameter length.

📋 What You'll Learn

Create a tree using adjacency lists

Use a helper variable to track the diameter

Implement a depth-first search (DFS) with dynamic programming to find the diameter

Print the diameter length

💡 Why This Matters

🌍 Real World

Finding the diameter of a tree is useful in network design, biology (evolution trees), and computer graphics (scene graphs).

💼 Career

Understanding tree diameters and dynamic programming on trees is important for software engineers working on graph algorithms, optimization problems, and system design.

1

Create the tree using adjacency lists

Create a dictionary called tree where keys are integers representing nodes and values are lists of integers representing connected nodes. Add these exact edges: 1 connected to 2 and 3, 2 connected to 4 and 5.

DSA C#

# Your code here

Hint

Use a Dictionary<int, List<int>> to represent the tree. Each key is a node, and the list contains its connected nodes.

2

Add a helper variable to track the diameter

Add an integer variable called diameter and set it to 0 before the DFS function.

DSA C#

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        Dictionary> tree = new Dictionary>
        {
            {1, new List{2, 3}},
            {2, new List{1, 4, 5}},
            {3, new List{1}},
            {4, new List{2}},
            {5, new List{2}}
        };

        // Add diameter variable here
        // Your code here
    }
}

Hint

Declare diameter as a static integer variable outside Main so it can be updated inside the DFS function.

3

Implement DFS to find the diameter

Add a static method called DFS that takes int node, int parent, and Dictionary<int, List<int>> tree. Inside DFS, find the two longest depths from children, update diameter with their sum if larger, and return the longest depth plus one. Call DFS(1, -1, tree) inside Main.

DSA C#

    static int DFS(int node, int parent, Dictionary> tree)
        int max1 = 0, max2 = 0;
        foreach (int child in tree[node])
            if (child == parent) continue;
            int depth = DFS(child, node, tree);
            if (depth > max1)
            {
                max2 = max1;
                max1 = depth;
            }
            else if (depth > max2)
            {
                max2 = depth;
            }
        }
        diameter = Math.Max(diameter, max1 + max2);
        return max1 + 1;
        DFS(1, -1, tree);

Hint

Use DFS to explore children. Track the two longest depths to update the diameter. Return the longest depth plus one for the parent.

4

Print the diameter length

Add a Console.WriteLine(diameter); statement at the end of Main to print the diameter length.

DSA C#

Hint

The diameter of this tree is 3. It is the longest path between nodes 4 and 3 or 5 and 3.

Practice

1. What does the diameter of a tree represent in the context of DP on Trees?

easy

A. The height of the tree from the root

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

C. The number of nodes in the tree

D. The total sum of all node values

Solution

1. Step 1: Understand the definition of tree diameter

   The diameter is the longest path between any two nodes in a tree, not necessarily passing through the root.

2. Step 2: Differentiate diameter from other tree properties

   Height is the longest path from root to leaf, number of nodes is size, and sum of values is unrelated to diameter.

3. Final Answer:

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

4. Quick Check:

   Diameter = longest path [OK]

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

Common Mistakes:

• Confusing diameter with height
• Thinking diameter is number of nodes
• Assuming diameter passes through root

2. Which of the following C# code snippets correctly initializes an adjacency list for a tree with n nodes?

easy

A. var graph = new List<int>[n]; for(int i=0; i

B. var graph = new List<int>(n); for(int i=0; i

C. var graph = new List(); for(int i=0; i

D. var graph = new List<int>[n]; graph = new List<int>();

Solution

1. Step 1: Understand adjacency list initialization in C#

   For n nodes, create an array of List<int> with size n, then initialize each element with new List<int>().

2. Step 2: Analyze each option

   A: var graph = new List<int>[n]; for(int i=0; i(); for(int i=0; i

3. Final Answer:

   var graph = new List<int>[n]; for(int i=0; i -> Option A

4. Quick Check:

   Array of lists initialized correctly [OK]

Hint: Use array of lists and initialize each element [OK]

Common Mistakes:

• Not initializing each list element
• Using List<int> without specifying size
• Overwriting array with a single list

3. Consider the following C# code snippet to compute the diameter of a tree. What will be the output printed?

using System;
using System.Collections.Generic;

class Program {
    static List<int>[] graph;
    static int diameter = 0;

    static int DFS(int node, int parent) {
        int max1 = 0, max2 = 0;
        foreach (var child in graph[node]) {
            if (child == parent) continue;
            int dist = DFS(child, node) + 1;
            if (dist > max1) {
                max2 = max1;
                max1 = dist;
            } else if (dist > max2) {
                max2 = dist;
            }
        }
        diameter = Math.Max(diameter, max1 + max2);
        return max1;
    }

    static void Main() {
        int n = 5;
        graph = new List<int>[n];
        for (int i = 0; i < n; i++) graph[i] = new List<int>();
        graph[0].Add(1); graph[1].Add(0);
        graph[1].Add(2); graph[2].Add(1);
        graph[1].Add(3); graph[3].Add(1);
        graph[3].Add(4); graph[4].Add(3);

        DFS(0, -1);
        Console.WriteLine(diameter);
    }
}

medium

A. 2

B. 5

C. 3

D. 4

Solution

1. Step 1: Understand the tree structure and DFS traversal

   The tree edges are: 0-1, 1-2, 1-3, 3-4. The longest path is from node 2 to node 4 passing through 1 and 3, length 3 edges.

2. Step 2: Trace DFS and diameter calculation

   DFS calculates max distances from each node's children. Diameter updates to max sum of two longest child paths. The longest path length is 3.

3. Final Answer:

   3 -> Option C

4. Quick Check:

   Longest path length = 3 edges [OK]

Hint: Diameter = longest path edges between two nodes [OK]

Common Mistakes:

• Counting nodes instead of edges
• Misunderstanding DFS return values
• Ignoring parent check in DFS

4. The following code attempts to compute the diameter of a tree but produces incorrect results. What is the main bug?

static int DFS(int node, int parent) {
    int maxDist = 0;
    foreach (var child in graph[node]) {
        if (child == parent) continue;
        int dist = DFS(child, node) + 1;
        if (dist > maxDist) maxDist = dist;
    }
    diameter = Math.Max(diameter, maxDist);
    return maxDist;
}

medium

A. It does not check for parent node, causing infinite recursion

B. It updates diameter inside the loop incorrectly

C. It returns distance without adding 1 for edge

D. It only considers the longest single path, missing combined paths for diameter

Solution

1. Step 1: Analyze how diameter is updated

   The code tracks only the longest single path from children, ignoring the sum of two longest paths needed for diameter.

2. Step 2: Understand diameter definition

   Diameter is the longest path between any two nodes, which may be sum of two longest child paths through current node, not just one.

3. Final Answer:

   It only considers the longest single path, missing combined paths for diameter -> Option D

4. Quick Check:

   Diameter needs sum of two longest paths [OK]

Hint: Diameter = sum of two longest child paths, not just one [OK]

Common Mistakes:

• Ignoring second longest path
• Forgetting to skip parent node
• Updating diameter inside loop incorrectly

5. Given a weighted tree where edges have positive weights, how would you modify the DP approach to find the diameter of the tree?

hard

A. Replace increments by edge weights when calculating distances in DFS

B. Ignore weights and count edges as 1 each

C. Use BFS instead of DFS without changes

D. Sum all edge weights and divide by number of edges

Solution

1. Step 1: Understand difference between unweighted and weighted trees

   In weighted trees, edge lengths vary, so distance must accumulate actual weights, not just count edges.

2. Step 2: Modify DFS to add edge weights

   Instead of adding 1 for each edge, add the weight of the edge when calculating distances and updating diameter.

3. Step 3: Ignore incorrect options

   Ignoring weights loses accuracy, BFS alone doesn't solve weighted distances, summing all weights is unrelated to diameter.

4. Final Answer:

   Replace increments by edge weights when calculating distances in DFS -> Option A

5. Quick Check:

   Use edge weights in distance sums [OK]

Hint: Add edge weights instead of 1 in DFS distance calculation [OK]

Common Mistakes:

• Counting edges as 1 ignoring weights
• Using BFS without weight handling
• Summing all weights unrelated to diameter