DFS Depth First Search on Graph in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

DFS traversal order on a simple graph

What is the output order of nodes visited by the DFS starting from node 0?

DSA C#

using System;
using System.Collections.Generic;

class Program {
    static Dictionary<int, List<int>> graph = new Dictionary<int, List<int>> {
        {0, new List<int>{1, 2}},
        {1, new List<int>{3}},
        {2, new List<int>{4}},
        {3, new List<int>()},
        {4, new List<int>()}
    };

    static HashSet<int> visited = new HashSet<int>();

    static void DFS(int node) {
        Console.Write(node + " ");
        visited.Add(node);
        foreach (var neighbor in graph[node]) {
            if (!visited.Contains(neighbor)) {
                DFS(neighbor);
            }
        }
    }

    static void Main() {
        DFS(0);
    }
}

A 0 1 3 2 4

B 0 2 4 1 3

C 0 1 2 3 4

D 0 3 1 4 2

💡 Hint

Remember DFS explores as far as possible along each branch before backtracking.

Starting at node 0, DFS visits 0, then goes to 1, then 3 (no neighbors), backtracks to 0, then visits 2, then 4.

❓ Predict Output

intermediate

DFS visit count on disconnected graph

How many nodes will be visited by DFS starting from node 1 in this graph?

DSA C#

using System;
using System.Collections.Generic;

class Program {
    static Dictionary<int, List<int>> graph = new Dictionary<int, List<int>> {
        {0, new List<int>{1}},
        {1, new List<int>{2}},
        {2, new List<int>()},
        {3, new List<int>{4}},
        {4, new List<int>()}
    };

    static HashSet<int> visited = new HashSet<int>();

    static void DFS(int node) {
        visited.Add(node);
        foreach (var neighbor in graph[node]) {
            if (!visited.Contains(neighbor)) {
                DFS(neighbor);
            }
        }
    }

    static void Main() {
        DFS(1);
        Console.WriteLine(visited.Count);
    }
}

A 3

B 2

C 4

D 5

💡 Hint

DFS only visits nodes reachable from the start node.

Starting at node 1, DFS visits nodes 1 and 2. Node 0 is not visited because DFS does not go backwards. So total visited nodes are 2.

🔧 Debug

advanced

Identify the error in DFS implementation

What error will this DFS code produce when run?

DSA C#

using System;
using System.Collections.Generic;

class Program {
    static Dictionary<int, List<int>> graph = new Dictionary<int, List<int>> {
        {0, new List<int>{1}},
        {1, new List<int>{2}},
        {2, new List<int>()}
    };

    static HashSet<int> visited = new HashSet<int>();

    static void DFS(int node) {
        if (visited.Contains(node)) return;
        foreach (var neighbor in graph[node]) {
            DFS(neighbor);
        }
        visited.Add(node);
    }

    static void Main() {
        DFS(0);
        Console.WriteLine(string.Join(",", visited));
    }
}

A Runtime error: KeyNotFoundException

B Output: 0,2,1

C Output: 0,1,2

D Output: 2,1,0

💡 Hint

Check when nodes are marked visited in DFS.

Nodes are added to visited after visiting neighbors, so the output order is post-order: 2,1,0.

❓ Predict Output

advanced

DFS with cycle detection output

What will be printed by this DFS code that detects cycles?

DSA C#

using System;
using System.Collections.Generic;

class Program {
    static Dictionary<int, List<int>> graph = new Dictionary<int, List<int>> {
        {0, new List<int>{1}},
        {1, new List<int>{2}},
        {2, new List<int>{0}}
    };

    static HashSet<int> visited = new HashSet<int>();
    static HashSet<int> recursionStack = new HashSet<int>();
    static bool hasCycle = false;

    static void DFS(int node) {
        if (recursionStack.Contains(node)) {
            hasCycle = true;
            return;
        }
        if (visited.Contains(node)) return;

        visited.Add(node);
        recursionStack.Add(node);

        foreach (var neighbor in graph[node]) {
            DFS(neighbor);
        }

        recursionStack.Remove(node);
    }

    static void Main() {
        DFS(0);
        Console.WriteLine(hasCycle);
    }
}

A True

B False

C Runtime error: StackOverflowException

D Compilation error

💡 Hint

A cycle exists if a node is revisited in the current recursion path.

The graph has a cycle 0->1->2->0, so hasCycle becomes true and prints True.

🧠 Conceptual

expert

DFS stack vs recursion behavior

Which statement about DFS implementation using explicit stack vs recursion is correct?

A Stack-based DFS cannot detect cycles, but recursive DFS can.

B Recursive DFS always visits nodes in a different order than stack-based DFS.

C Both methods visit nodes in the same order if neighbors are pushed in the same order.

D Recursive DFS uses more memory than stack-based DFS because it stores all nodes in the call stack.

💡 Hint

Think about how the order of neighbor processing affects traversal.

Both recursive and stack-based DFS visit nodes in the same order if neighbors are processed in the same order because the stack simulates recursion.

Practice

1. What is the main idea behind Depth First Search (DFS) on a graph?

easy

A. Visit nodes in order of their distance from the start node

B. Visit all neighbors of a node before moving deeper

C. Explore as far as possible along each branch before backtracking

D. Randomly visit nodes until all are visited

Solution

1. Step 1: Understand DFS traversal

   DFS explores a graph by going deep into one branch before backtracking to explore others.

2. Step 2: Compare with other traversal methods

   BFS visits neighbors first, but DFS goes deep first.

3. Final Answer:

   Explore as far as possible along each branch before backtracking -> Option C

4. Quick Check:

   DFS = deep exploration first [OK]

Hint: DFS means go deep first, then backtrack [OK]

Common Mistakes:

• Confusing DFS with BFS
• Thinking DFS visits neighbors first
• Assuming DFS visits nodes by distance

2. Which of the following is the correct way to start a DFS traversal in C# using recursion on a graph represented by adjacency list Dictionary<int, List<int>> graph?

easy

A. void DFS(int node, HashSet<int> visited) { if (visited.Contains(node)) return; visited.Add(node); foreach (var neighbor in graph[node]) DFS(neighbor, visited); }

B. void DFS(int node) { visited.Add(node); foreach (var neighbor in graph[node]) DFS(neighbor); }

C. void DFS(int node, List<int> visited) { if (visited.Contains(node)) return; visited.Add(node); foreach (var neighbor in graph[node]) DFS(neighbor, visited); }

D. void DFS(int node, HashSet<int> visited) { visited.Add(node); foreach (var neighbor in graph) DFS(neighbor, visited); }

Solution

1. Step 1: Check parameters and visited tracking

   void DFS(int node, HashSet<int> visited) { if (visited.Contains(node)) return; visited.Add(node); foreach (var neighbor in graph[node]) DFS(neighbor, visited); } uses HashSet to track visited nodes and prevents revisiting.

2. Step 2: Verify iteration over neighbors

   void DFS(int node, HashSet<int> visited) { if (visited.Contains(node)) return; visited.Add(node); foreach (var neighbor in graph[node]) DFS(neighbor, visited); } correctly iterates over neighbors of the current node from the graph dictionary.

3. Final Answer:

   void DFS(int node, HashSet<int> visited) { if (visited.Contains(node)) return; visited.Add(node); foreach (var neighbor in graph[node]) DFS(neighbor, visited); } -> Option A

4. Quick Check:

   Correct recursion with visited check [OK]

Hint: Use HashSet to track visited nodes in DFS recursion [OK]

Common Mistakes:

• Not checking if node is visited before recursion
• Using List instead of HashSet for visited (slow lookup)
• Iterating over graph keys instead of neighbors

3. Given the graph adjacency list:

var graph = new Dictionary {
  {1, new List<int>{2, 3}},
  {2, new List<int>{4}},
  {3, new List<int>{4}},
  {4, new List<int>()}
};

List<int> visited = new List<int>();
DFS(1, visited); // DFS as defined in option C above
Console.WriteLine(string.Join(", ", visited));

What is the output?

medium

A. 1, 2, 3, 4

B. 1, 2, 4, 3

C. 1, 3, 4, 2

D. 1, 4, 2, 3

Solution

1. Step 1: Start DFS from node 1

   Visit 1, mark visited. Neighbors: 2, 3.

2. Step 2: Visit neighbor 2 first

   Visit 2, then its neighbor 4. 4 has no neighbors, backtrack.

3. Step 3: Backtrack to 1 and visit neighbor 3

   Visit 3, then neighbor 4 (already visited, skip).

4. Step 4: Order of visited nodes

   Visited nodes in order: 1, 2, 4, 3.

5. Final Answer:

   1, 2, 4, 3 -> Option B

6. Quick Check:

   DFS order = 1,2,4,3 [OK]

Hint: DFS visits neighbors in order, backtracks after deepest node [OK]

Common Mistakes:

• Assuming BFS order instead of DFS
• Visiting node 3 before node 2
• Revisiting already visited nodes

4. Consider this DFS code snippet in C#:

void DFS(int node, HashSet<int> visited) {
  visited.Add(node);
  foreach (var neighbor in graph.Keys) {
    if (!visited.Contains(neighbor)) {
      DFS(neighbor, visited);
    }
  }
}

What is the main problem with this DFS implementation?

medium

A. It iterates over all graph nodes instead of neighbors of the current node

B. It does not add the current node to visited

C. It uses List instead of HashSet for visited nodes

D. It does not check if the node is null before recursion

Solution

1. Step 1: Analyze the foreach loop

   The loop iterates over all keys in the graph, not neighbors of the current node.

2. Step 2: Understand DFS neighbor traversal

   DFS should explore neighbors of the current node, not all nodes in the graph.

3. Final Answer:

   It iterates over all graph nodes instead of neighbors of the current node -> Option A

4. Quick Check:

   DFS must visit neighbors, not all nodes [OK]

Hint: Loop over neighbors, not all graph keys [OK]

Common Mistakes:

• Looping over graph.Keys instead of graph[node]
• Forgetting to add node to visited
• Not checking visited before recursion

5. You want to find if there is a path from node 1 to node 5 in this directed graph:

var graph = new Dictionary {
  {1, new List<int>{2, 3}},
  {2, new List<int>{4}},
  {3, new List<int>{4, 5}},
  {4, new List<int>{}},
  {5, new List<int>{}}
};

bool DFS(int current, int target, HashSet<int> visited) {
  if (current == target) return true;
  visited.Add(current);
  foreach (var neighbor in graph[current]) {
    if (!visited.Contains(neighbor)) {
      if (DFS(neighbor, target, visited)) return true;
    }
  }
  return false;
}

var visited = new HashSet<int>();
bool result = DFS(1, 5, visited);
Console.WriteLine(result);

What will be printed?

hard

A. Runtime exception

B. False

C. Compilation error

D. True

Solution

1. Step 1: Start DFS from node 1 looking for 5

   Visit 1, neighbors 2 and 3.

2. Step 2: Explore neighbor 2

   From 2, visit 4. 4 has no neighbors, backtrack.

3. Step 3: Explore neighbor 3

   From 3, neighbors are 4 and 5. 4 visited, then 5 found matches target.

4. Step 4: DFS returns true when target found

   DFS returns true and prints True.

5. Final Answer:

   True -> Option D

6. Quick Check:

   Path exists from 1 to 5 [OK]

Hint: DFS returns true immediately when target found [OK]

Common Mistakes:

• Assuming no path because 5 is not direct neighbor
• Not returning true when target found
• Forgetting to mark visited nodes