DP on Trees Diameter of Tree in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Diameter Calculation on a Simple Tree

What is the output of the following C# code that calculates the diameter of a tree represented as an adjacency list?

DSA C#

using System;
using System.Collections.Generic;

class Program {
    static List<int>[] tree;
    static int maxDiameter = 0;

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

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

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

A 3

B 5

C 4

D 2

💡 Hint

Trace the longest path between any two nodes in the tree.

The longest path (diameter) in this tree is from node 2 to node 4 passing through nodes 1 and 3, which has length 3 edges.

🧠 Conceptual

intermediate

Understanding Diameter Calculation Logic

In the diameter calculation algorithm using DFS on a tree, why do we keep track of the two longest depths from children of a node?

A Because the diameter may pass through the current node connecting two longest child paths.

B Because we need to find the shortest path between two nodes.

C Because the diameter is always the depth of the deepest child.

D Because the sum of all child depths gives the diameter.

💡 Hint

Think about the longest path that goes through the current node.

The diameter can be the longest path passing through the current node, which is the sum of the two longest depths from its children.

❓ Predict Output

advanced

Output of Diameter Calculation on a Star-Shaped Tree

What is the output of the following C# code that calculates the diameter of a star-shaped tree with 4 leaves connected to a center node?

DSA C#

using System;
using System.Collections.Generic;

class Program {
    static List<int>[] tree;
    static int maxDiameter = 0;

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

    static void Main() {
        int n = 5;
        tree = new List<int>[n];
        for (int i = 0; i < n; i++) tree[i] = new List<int>();
        // star shape: node 0 connected to 1,2,3,4
        for (int i = 1; i < n; i++) {
            tree[0].Add(i);
            tree[i].Add(0);
        }

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

A 1

B 3

C 4

D 2

💡 Hint

The longest path in a star is between any two leaves through the center.

The diameter is 2 edges: from one leaf to center to another leaf.

🔧 Debug

advanced

Identify the Error in Diameter Calculation Code

What error will occur when running the following C# code to calculate the diameter of a tree?

DSA C#

using System;
using System.Collections.Generic;

class Program {
    static List<int>[] tree;
    static int maxDiameter = 0;

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

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

        DFS(3, -1);
        Console.WriteLine(maxDiameter);
    }
}

A StackOverflowException

B No error, outputs 2

C IndexOutOfRangeException

D NullReferenceException

💡 Hint

Check the node index passed to DFS in Main.

DFS is called with node 3 but tree size is 3 (indices 0 to 2), causing IndexOutOfRangeException.

🚀 Application

expert

Calculate Diameter for a Complex Tree

Given the following tree edges, what is the diameter of the tree? Edges: 0-1, 1-2, 1-3, 3-4, 4-5, 5-6 Use the diameter definition as the longest path between any two nodes.

A 5

B 6

C 7

D 4

💡 Hint

Find the longest path by tracing from leaf to leaf through the tree.

The longest path is from node 2 to node 6 passing through nodes 1,3,4,5 with length 6 edges.

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