Print all paths from a given source to a destination using BFS

Given a directed graph, a source vertex ‘src’ and a destination vertex ‘dst’, print all paths from given ‘src’ to ‘dst’.

Consider the following directed graph. Let the src be 2 and dst be 3. There are 3 different paths from 2 to 3.

We have already discussed Print all paths from a given source to a destination using DFS.

Below is BFS based solution.

Algorithm :

create a queue which will store path(s) of type vector
initialise the queue with first path starting from src

Now run a loop till queue is not empty
   get the frontmost path from queue
   check if the lastnode of this path is destination
       if true then print the path
   run a loop for all the vertices connected to the
   current vertex i.e. lastnode extracted from path
      if the vertex is not visited in current path
         a) create a new path from earlier path and 
             append this vertex
         b) insert this new path to queue

Implementation:

C++

// C++ program to print all paths of source to
// destination in given graph
#include <bits/stdc++.h>
using namespace std;

// utility function for printing
// the found path in graph
void printpath(vector<int>& path)
{
    int size = path.size();
    for (int i = 0; i < size; i++)
        cout << path[i] << " ";
    cout << endl;
}

// utility function to check if current
// vertex is already present in path
int isNotVisited(int x, vector<int>& path)
{
    int size = path.size();
    for (int i = 0; i < size; i++)
        if (path[i] == x)
            return 0;
    return 1;
}

// utility function for finding paths in graph
// from source to destination
void findpaths(vector<vector<int> >& g, int src, int dst,
               int v)
{
    // create a queue which stores
    // the paths
    queue<vector<int> > q;

    // path vector to store the current path
    vector<int> path;
    path.push_back(src);
    q.push(path);
    while (!q.empty()) {
        path = q.front();
        q.pop();
        int last = path[path.size() - 1];

        // if last vertex is the desired destination
        // then print the path
        if (last == dst)
            printpath(path);

        // traverse to all the nodes connected to
        // current vertex and push new path to queue
        for (int i = 0; i < g[last].size(); i++) {
            if (isNotVisited(g[last][i], path)) {
                vector<int> newpath(path);
                newpath.push_back(g[last][i]);
                q.push(newpath);
            }
        }
    }
}

// driver program
int main()
{
    vector<vector<int> > g;
    // number of vertices
    int v = 4;
    g.resize(4);

    // construct a graph
    g[0].push_back(3);
    g[0].push_back(1);
    g[0].push_back(2);
    g[1].push_back(3);
    g[2].push_back(0);
    g[2].push_back(1);

    int src = 2, dst = 3;
    cout << "path from src " << src << " to dst " << dst
         << " are \n";

    // function for finding the paths
    findpaths(g, src, dst, v);

    return 0;
}

Java

// Java program to print all paths of source to
// destination in given graph
import java.io.*;
import java.util.*;

class Graph{

// utility function for printing
// the found path in graph
private static void printPath(List<Integer> path)
{
    int size = path.size();
    for(Integer v : path)
    {
        System.out.print(v + " ");
    }
    System.out.println();
}

// Utility function to check if current
// vertex is already present in path
private static boolean isNotVisited(int x,
                                    List<Integer> path)
{
    int size = path.size();
    for(int i = 0; i < size; i++)
        if (path.get(i) == x)
            return false;

    return true;
}

// Utility function for finding paths in graph
// from source to destination
private static void findpaths(List<List<Integer> > g,
                              int src, int dst, int v)
{

    // Create a queue which stores
    // the paths
    Queue<List<Integer> > queue = new LinkedList<>();

    // Path vector to store the current path
    List<Integer> path = new ArrayList<>();
    path.add(src);
    queue.offer(path);

    while (!queue.isEmpty())
    {
        path = queue.poll();
        int last = path.get(path.size() - 1);

        // If last vertex is the desired destination
        // then print the path
        if (last == dst)
        {
            printPath(path);
        }

        // Traverse to all the nodes connected to
        // current vertex and push new path to queue
        List<Integer> lastNode = g.get(last);
        for(int i = 0; i < lastNode.size(); i++)
        {
            if (isNotVisited(lastNode.get(i), path))
            {
                List<Integer> newpath = new ArrayList<>(path);
                newpath.add(lastNode.get(i));
                queue.offer(newpath);
            }
        }
    }
}

// Driver code
public static void main(String[] args)
{
    List<List<Integer> > g = new ArrayList<>();
    int v = 4;
    for(int i = 0; i < 4; i++)
    {
        g.add(new ArrayList<>());
    }

    // Construct a graph
    g.get(0).add(3);
    g.get(0).add(1);
    g.get(0).add(2);
    g.get(1).add(3);
    g.get(2).add(0);
    g.get(2).add(1);
    int src = 2, dst = 3;
    System.out.println("path from src " + src +
                       " to dst " + dst + " are ");

    // Function for finding the paths
    findpaths(g, src, dst, v);
}
}

// This code is contributed by rajatsri94

Python3

# Python3 program to print all paths of
# source to destination in given graph
from typing import List
from collections import deque

# Utility function for printing
# the found path in graph
def printpath(path: List[int]) -> None:

    size = len(path)
    for i in range(size):
        print(path[i], end = " ")

    print()

# Utility function to check if current
# vertex is already present in path
def isNotVisited(x: int, path: List[int]) -> int:

    size = len(path)
    for i in range(size):
        if (path[i] == x):
            return 0

    return 1

# Utility function for finding paths in graph
# from source to destination
def findpaths(g: List[List[int]], src: int,
              dst: int, v: int) -> None:

    # Create a queue which stores
    # the paths
    q = deque()

    # Path vector to store the current path
    path = []
    path.append(src)
    q.append(path.copy())

    while q:
        path = q.popleft()
        last = path[len(path) - 1]

        # If last vertex is the desired destination
        # then print the path
        if (last == dst):
            printpath(path)

        # Traverse to all the nodes connected to
        # current vertex and push new path to queue
        for i in range(len(g[last])):
            if (isNotVisited(g[last][i], path)):
                newpath = path.copy()
                newpath.append(g[last][i])
                q.append(newpath)

# Driver code
if __name__ == "__main__":

    # Number of vertices
    v = 4
    g = [[] for _ in range(4)]

    # Construct a graph
    g[0].append(3)
    g[0].append(1)
    g[0].append(2)
    g[1].append(3)
    g[2].append(0)
    g[2].append(1)

    src = 2
    dst = 3
    print("path from src {} to dst {} are".format(
        src, dst))

    # Function for finding the paths
    findpaths(g, src, dst, v)

# This code is contributed by sanjeev2552

C#

// C# program to print all paths of source to
// destination in given graph

using System;
using System.Collections.Generic;

public class Graph{

// utility function for printing
// the found path in graph
static void printPath(List<int> path)
{
    int size = path.Count;
    foreach(int v in path)
    {
        Console.Write(v + " ");
    }
    Console.WriteLine();
}

// Utility function to check if current
// vertex is already present in path
static bool isNotVisited(int x, List<int> path)
{
    int size = path.Count;
    for(int i = 0; i < size; i++)
        if (path[i] == x)
            return false;

    return true;
}

// Utility function for finding paths in graph
// from source to destination
private static void findpaths(List<List<int> > g,
                              int src, int dst, int v)
{

    // Create a queue which stores
    // the paths
    Queue<List<int> > queue = new Queue<List<int>>();

    // Path vector to store the current path
    List<int> path = new List<int>();
    path.Add(src);
    queue.Enqueue(path);

    while (queue.Count!=0)
    {
        path = queue.Dequeue();
        int last = path[path.Count - 1];

        // If last vertex is the desired destination
        // then print the path
        if (last == dst)
        {
            printPath(path);
        }

        // Traverse to all the nodes connected to
        // current vertex and push new path to queue
        List<int> lastNode = g[last];
        for(int i = 0; i < lastNode.Count; i++)
        {
            if (isNotVisited(lastNode[i], path))
            {
                List<int> newpath = new List<int>(path);
                newpath.Add(lastNode[i]);
                queue.Enqueue(newpath);
            }
        }
    }
}

// Driver code
public static void Main(String[] args)
{
    List<List<int> > g = new List<List<int>>();
    int v = 4;
    for(int i = 0; i < 4; i++)
    {
        g.Add(new List<int>());
    }

    // Construct a graph
    g[0].Add(3);
    g[0].Add(1);
    g[0].Add(2);
    g[1].Add(3);
    g[2].Add(0);
    g[2].Add(1);
    int src = 2, dst = 3;
    Console.WriteLine("path from src " + src +
                       " to dst " + dst + " are ");

    // Function for finding the paths
    findpaths(g, src, dst, v);
}
}

// This code is contributed by shikhasingrajput

Javascript

// JavaScript code to print all paths of
// source to destination in given graph

// Utility function for printing
// the found path in graph
function printpath(path) {
    let size = path.length;
    for (let i = 0; i < size; i++) {
        process.stdout.write(path[i] + " ");
    }
    console.log();
}

// Utility function to check if current
// vertex is already present in path
function isNotVisited(x, path) {
    let size = path.length;
    for (let i = 0; i < size; i++) {
        if (path[i] === x) {
            return 0;
        }
    }
    return 1;
}

// Utility function for finding paths in graph
// from source to destination
function findpaths(g, src, dst, v) {
    // Create a queue which stores
    // the paths
    let q = [];

    // Path array to store the current path
    let path = [];
    path.push(src);
    q.push(path.slice());

    while (q.length) {
        path = q.shift();
        let last = path[path.length - 1];

        // If last vertex is the desired destination
        // then print the path
        if (last === dst) {
            printpath(path);
        }

        // Traverse to all the nodes connected to
        // current vertex and push new path to queue
        for (let i = 0; i < g[last].length; i++) {
            if (isNotVisited(g[last][i], path)) {
                let newpath = path.slice();
                newpath.push(g[last][i]);
                q.push(newpath);
            }
        }
    }
}

// Driver code

    // Number of vertices
    let v = 4;
    let g = Array.from({ length: 4 }, () => []);

    // Construct a graph
    g[0].push(3);
    g[0].push(1);
    g[0].push(2);
    g[1].push(3);
    g[2].push(0);
    g[2].push(1);

    let src = 2;
    let dst = 3;
    console.log(`path from src ${src} to dst ${dst} are`);

    // Function for finding the paths
    findpaths(g, src, dst, v);

Output

path from src 2 to dst 3 are 
2 0 3 
2 1 3 
2 0 1 3 

Time Complexity: O(V^V), the time complexity is exponential. From each vertex there are v vertices that can be visited from current vertex.
Auxiliary space: O(V^V), as there can be V^V paths possible in the worst case.

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