BFS Breadth First Search on Graph in DSA C++ - Build from Scratch

BFS Breadth First Search on Graph

📖 Scenario: You are working with a simple map of connected cities. Each city is connected to some other cities by roads. You want to visit all cities starting from a specific city, exploring neighbors first before going deeper.

🎯 Goal: Build a program that uses BFS (Breadth First Search) to visit all cities starting from city 0 and print the order in which cities are visited.

📋 What You'll Learn

Create an adjacency list to represent the graph of 5 cities with given connections

Create a queue to help with BFS traversal

Implement BFS traversal starting from city 0

Print the order of cities visited by BFS

💡 Why This Matters

🌍 Real World

BFS is used in social networks to find friends at different levels, in GPS to find shortest paths, and in web crawlers to explore websites layer by layer.

💼 Career

Understanding BFS is essential for software engineers working on graph problems, network analysis, and algorithms for search and traversal.

1

Create the graph adjacency list

Create a vector of vectors called graph with 5 elements. Add these edges exactly: city 0 connected to 1 and 2, city 1 connected to 2 and 3, city 2 connected to 4, city 3 connected to 4.

DSA C++

std::vector> graph(5);
// Your code here

Hint

Use a vector of vectors to store neighbors for each city. Assign neighbors exactly as given.

2

Create a queue and visited list

Create a std::queue<int> called q and a std::vector<bool> called visited with 5 elements initialized to false. Then push city 0 into q and mark visited[0] as true.

DSA C++

std::queue q;
std::vector visited(5, false);
// Your code here

Hint

Use a queue to keep track of cities to visit next. Use a boolean vector to mark visited cities.

3

Implement BFS traversal loop

Write a while loop that runs while q is not empty. Inside, pop the front city into current. For each neighbor neighbor in graph[current], if visited[neighbor] is false, push neighbor into q and mark visited[neighbor] as true. Also, add each visited city to a vector order in the order they are visited. Initialize order with city 0 before the loop.

DSA C++

std::vector order = {0};
// Your code here

Hint

Use a while loop to process cities in the queue. For each city, add unvisited neighbors to the queue and mark them visited.

4

Print the BFS visit order

Print the cities in order separated by spaces on one line.

DSA C++

for (int city : order) {
    // Your code here
}

Hint

Use a for loop to print each city in order separated by spaces.

Practice

1. What does BFS (Breadth First Search) primarily use to keep track of nodes to visit next in a graph?

easy

A. A queue data structure

B. A stack data structure

C. A priority queue

D. A hash map

Solution

1. Step 1: Understand BFS traversal method

   BFS explores nodes level by level, visiting all neighbors before moving deeper.

2. Step 2: Identify the data structure used

   To maintain the order of visiting nodes level-wise, BFS uses a queue which follows FIFO (First In First Out).

3. Final Answer:

   A queue data structure -> Option A

4. Quick Check:

   BFS uses queue = B [OK]

Hint: BFS uses FIFO order, so it needs a queue [OK]

Common Mistakes:

• Confusing BFS with DFS which uses a stack
• Thinking BFS uses a priority queue
• Assuming BFS uses a hash map for order

2. Which of the following is the correct way to declare a queue for BFS traversal of integers in C++?

easy

A. vector<int> q;

B. queue<int> q;

C. stack<int> q;

D. list<int> q;

Solution

1. Step 1: Identify the correct STL container for queue

   In C++, the standard queue container is declared as queue<type>.

2. Step 2: Match the declaration for integer queue

   queue<int> q; correctly declares a queue of integers.

3. Final Answer:

   queue<int> q; -> Option B

4. Quick Check:

   Correct queue syntax = queue<int> q; [OK]

Hint: Use queue<int> for BFS integer queue in C++ [OK]

Common Mistakes:

• Using stack instead of queue
• Using vector or list which are not queues
• Missing template angle brackets

3. Consider the following graph represented as adjacency list:

0: 1, 2
1: 0, 3
2: 0, 3
3: 1, 2

If BFS starts from node 0, what is the order of nodes visited?

medium

A. [0, 1, 3, 2]

B. [0, 2, 1, 3]

C. [0, 3, 1, 2]

D. [0, 1, 2, 3]

Solution

1. Step 1: Start BFS from node 0

   Visit node 0 first, enqueue neighbors 1 and 2.

2. Step 2: Visit neighbors in queue order

   Dequeue 1, enqueue its neighbor 3 (0 already visited). Then dequeue 2, enqueue 3 but already visited.

3. Step 3: Visit node 3

   Finally dequeue 3.

4. Final Answer:

   [0, 1, 2, 3] -> Option D

5. Quick Check:

   BFS order = [0, 1, 2, 3] [OK]

Hint: Visit neighbors in queue order, avoid revisits [OK]

Common Mistakes:

• Visiting neighbors in wrong order
• Revisiting nodes already visited
• Confusing BFS with DFS order

4. The following BFS code snippet has a bug. What is the error?

void bfs(int start, vector& adj) {
  vector<bool> visited(adj.size(), false);
  queue<int> q;
  q.push(start);
  while (!q.empty()) {
    int node = q.front();
    q.pop();
    cout << node << " ";
    for (int neighbor : adj[node]) {
      if (!visited[neighbor]) {
        q.push(neighbor);
      }
    }
  }
}

medium

A. Neighbors are not iterated correctly

B. Queue is not initialized properly

C. Visited array is never updated to true for visited nodes

D. Start node is not pushed into the queue

Solution

1. Step 1: Check visited array usage

   Visited array is created but nodes are never marked visited after visiting.

2. Step 2: Understand impact on BFS

   Without marking visited true, nodes can be enqueued multiple times causing infinite loops or repeated visits.

3. Final Answer:

   Visited array is never updated to true for visited nodes -> Option C

4. Quick Check:

   Must mark visited[node] = true after popping [OK]

Hint: Always mark nodes visited after dequeuing [OK]

Common Mistakes:

• Forgetting to mark visited nodes
• Pushing start node multiple times
• Incorrect loop conditions

5. Given a graph with nodes 0 to 4 and edges: 0-1, 0-2, 1-3, 2-3, 3-4. Using BFS starting from node 0, which node is visited at the 3rd position in the BFS order?

hard

A. 2

B. 4

C. 1

D. 3

Solution

1. Step 1: List neighbors and BFS order

   Start at 0, enqueue neighbors 1 and 2. BFS order so far: [0]

2. Step 2: Visit nodes in queue order

   Dequeue 1 (2nd visited), enqueue 3. Dequeue 2 (3rd visited), enqueue 3 but already enqueued.

3. Step 3: Confirm 3rd visited node

   The third node visited is 2.

4. Final Answer:

   2 -> Option A

5. Quick Check:

   Third visited node = 2 [OK]

Hint: Track queue order carefully to find nth visited node [OK]

Common Mistakes:

• Mixing order of neighbors
• Counting start node incorrectly
• Ignoring repeated nodes in queue