Print all shortest paths between given source and destination in an undirected graph

Given an undirected and unweighted graph and two nodes as source and destination, the task is to print all the paths of the shortest length between the given source and destination.
Examples:

Input: source = 0, destination = 5 

Output:
0 -> 1 -> 3 -> 5
0 -> 2 -> 3 -> 5
0 -> 1 -> 4 -> 5 
Explanation:
All the above paths are of length 3, which is the shortest distance between 0 and 5.
Input: source = 0, destination = 4 

Output:
0 -> 1 -> 4 

Approach: The is to do a Breadth First Traversal (BFS) for a graph. Below are the steps:

1. Start BFS traversal from source vertex.
2. While doing BFS, store the shortest distance to each of the other nodes and also maintain a parent vector for each of the nodes.
3. Make the parent of source node as “-1”. For each node, it will store all the parents for which it has the shortest distance from the source node.
4. Recover all the paths using parent array. At any instant, we will push one vertex in the path array and then call for all its parents.
5. If we encounter “-1” in the above steps, then it means a path has been found and can be stored in the paths array.

Below is the implementation of the above approach:

cpp14

// Cpp program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to form edge between
// two vertices src and dest
void add_edge(vector<int> adj[],
              int src, int dest)
{
    adj[src].push_back(dest);
    adj[dest].push_back(src);
}

// Function which finds all the paths
// and stores it in paths array
void find_paths(vector<vector<int> >& paths,
                vector<int>& path,
                vector<int> parent[],
                int n, int u)
{
    // Base Case
    if (u == -1) {
        paths.push_back(path);
        return;
    }

    // Loop for all the parents
    // of the given vertex
    for (int par : parent[u]) {

        // Insert the current
        // vertex in path
        path.push_back(u);

        // Recursive call for its parent
        find_paths(paths, path, parent,
                   n, par);

        // Remove the current vertex
        path.pop_back();
    }
}

// Function which performs bfs
// from the given source vertex
void bfs(vector<int> adj[],
         vector<int> parent[],
         int n, int start)
{
    // dist will contain shortest distance
    // from start to every other vertex
    vector<int> dist(n, INT_MAX);

    queue<int> q;

    // Insert source vertex in queue and make
    // its parent -1 and distance 0
    q.push(start);
    parent[start] = { -1 };
    dist[start] = 0;

    // Until Queue is empty
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        for (int v : adj[u]) {
            if (dist[v] > dist[u] + 1) {

                // A shorter distance is found
                // So erase all the previous parents
                // and insert new parent u in parent[v]
                dist[v] = dist[u] + 1;
                q.push(v);
                parent[v].clear();
                parent[v].push_back(u);
            }
            else if (dist[v] == dist[u] + 1) {

                // Another candidate parent for
                // shortes path found
                parent[v].push_back(u);
            }
        }
    }
}

// Function which prints all the paths
// from start to end
void print_paths(vector<int> adj[],
                 int n, int start, int end)
{
    vector<vector<int> > paths;
    vector<int> path;
    vector<int> parent[n];

    // Function call to bfs
    bfs(adj, parent, n, start);

    // Function call to find_paths
    find_paths(paths, path, parent, n, end);

    for (auto v : paths) {

        // Since paths contain each
        // path in reverse order,
        // so reverse it
        reverse(v.begin(), v.end());

        // Print node for the current path
        for (int u : v)
            cout << u << " ";
        cout << endl;
    }
}

// Driver Code
int main()
{
    // Number of vertices
    int n = 6;

    // array of vectors is used
    // to store the graph
    // in the form of an adjacency list
    vector<int> adj[n];

    // Given Graph
    add_edge(adj, 0, 1);
    add_edge(adj, 0, 2);
    add_edge(adj, 1, 3);
    add_edge(adj, 1, 4);
    add_edge(adj, 2, 3);
    add_edge(adj, 3, 5);
    add_edge(adj, 4, 5);

    // Given source and destination
    int src = 0;
    int dest = n - 1;

    // Function Call
    print_paths(adj, n, src, dest);

    return 0;
}

Java

/*package whatever //do not write package name here */

import java.io.*;
import java.util.*;

class GFG {
    // Function to form edge between
    // two vertices src and dest
    static void add_edge(ArrayList<ArrayList<Integer>> adj, int src, int dest){
        adj.get(src).add(dest);
        adj.get(dest).add(src);
    }

    // Function which finds all the paths
    // and stores it in paths array
    static void find_paths(ArrayList<ArrayList<Integer>> paths, ArrayList<Integer> path,
                    ArrayList<ArrayList<Integer>> parent, int n, int u) {
        // Base Case
        if (u == -1) {
            paths.add(new ArrayList<>(path));
            return;
        }

        // Loop for all the parents
        // of the given vertex
        for (int par : parent.get(u)) {

            // Insert the current
            // vertex in path
            path.add(u);

            // Recursive call for its parent
            find_paths(paths, path, parent, n, par);

            // Remove the current vertex
            path.remove(path.size()-1);
        }
    }

    // Function which performs bfs
    // from the given source vertex
    static void bfs(ArrayList<ArrayList<Integer>> adj, ArrayList<ArrayList<Integer>> parent,
             int n, int start) {

        // dist will contain shortest distance
        // from start to every other vertex
          int[] dist = new int[n];
          Arrays.fill(dist, Integer.MAX_VALUE);

        Queue<Integer> q = new LinkedList<>();

        // Insert source vertex in queue and make
        // its parent -1 and distance 0
        q.offer(start);

        parent.get(start).clear();
          parent.get(start).add(-1);

        dist[start] = 0;

        // Until Queue is empty
        while (!q.isEmpty()) {
            int u = q.poll();

            for (int v : adj.get(u)) {
                if (dist[v] > dist[u] + 1) {

                    // A shorter distance is found
                    // So erase all the previous parents
                    // and insert new parent u in parent[v]
                    dist[v] = dist[u] + 1;
                    q.offer(v);
                    parent.get(v).clear();
                    parent.get(v).add(u);
                }
                else if (dist[v] == dist[u] + 1) {

                    // Another candidate parent for
                    // shortes path found
                    parent.get(v).add(u);
                }
            }
        }
    }

    // Function which prints all the paths
    // from start to end
    static void print_paths(ArrayList<ArrayList<Integer>> adj, int n, int start, int end){
        ArrayList<ArrayList<Integer>> paths = new ArrayList<>();
        ArrayList<Integer> path = new ArrayList<>();
        ArrayList<ArrayList<Integer>> parent = new ArrayList<>();

        for(int i = 0; i < n; i++){
            parent.add(new ArrayList<>());
        }

        // Function call to bfs
        bfs(adj, parent, n, start);

        // Function call to find_paths
        find_paths(paths, path, parent, n, end);

        for (ArrayList<Integer> v : paths) {

            // Since paths contain each
            // path in reverse order,
            // so reverse it
            Collections.reverse(v);

            // Print node for the current path
            for (int u : v)
                System.out.print(u + " ");

              System.out.println();
        }
    }

    public static void main (String[] args)
    {

      // Number of vertices
      int n = 6;

      // array of vectors is used
      // to store the graph
      // in the form of an adjacency list
      ArrayList<ArrayList<Integer>> adj = new ArrayList<>();
      for(int i = 0; i < n; i++){
          adj.add(new ArrayList<>());
      }

      // Given Graph
      add_edge(adj, 0, 1);
      add_edge(adj, 0, 2);
      add_edge(adj, 1, 3);
      add_edge(adj, 1, 4);
      add_edge(adj, 2, 3);
      add_edge(adj, 3, 5);
      add_edge(adj, 4, 5);

      // Given source and destination
      int src = 0;
      int dest = n - 1;

      // Function Call
      print_paths(adj, n, src, dest);

    }
}

// This code is contributed by ayush123ngp.

Python3

# Python program for the above approach

# Function to form edge between
# two vertices src and dest
from typing import List
from sys import maxsize
from collections import deque

def add_edge(adj: List[List[int]],
             src: int, dest: int) -> None:
    adj[src].append(dest)
    adj[dest].append(src)

# Function which finds all the paths
# and stores it in paths array
def find_paths(paths: List[List[int]], path: List[int],
               parent: List[List[int]], n: int, u: int) -> None:
    # Base Case
    if (u == -1):
        paths.append(path.copy())
        return

    # Loop for all the parents
    # of the given vertex
    for par in parent[u]:

        # Insert the current
        # vertex in path
        path.append(u)

        # Recursive call for its parent
        find_paths(paths, path, parent, n, par)

        # Remove the current vertex
        path.pop()

# Function which performs bfs
# from the given source vertex
def bfs(adj: List[List[int]],
        parent: List[List[int]], n: int,
        start: int) -> None:

    # dist will contain shortest distance
    # from start to every other vertex
    dist = [maxsize for _ in range(n)]
    q = deque()

    # Insert source vertex in queue and make
    # its parent -1 and distance 0
    q.append(start)
    parent[start] = [-1]
    dist[start] = 0

    # Until Queue is empty
    while q:
        u = q[0]
        q.popleft()
        for v in adj[u]:
            if (dist[v] > dist[u] + 1):

                # A shorter distance is found
                # So erase all the previous parents
                # and insert new parent u in parent[v]
                dist[v] = dist[u] + 1
                q.append(v)
                parent[v].clear()
                parent[v].append(u)

            elif (dist[v] == dist[u] + 1):

                # Another candidate parent for
                # shortes path found
                parent[v].append(u)

# Function which prints all the paths
# from start to end
def print_paths(adj: List[List[int]], n: int,
                start: int, end: int) -> None:
    paths = []
    path = []
    parent = [[] for _ in range(n)]

    # Function call to bfs
    bfs(adj, parent, n, start)

    # Function call to find_paths
    find_paths(paths, path, parent, n, end)
    for v in paths:

        # Since paths contain each
        # path in reverse order,
        # so reverse it
        v = reversed(v)

        # Print node for the current path
        for u in v:
            print(u, end = " ")
        print()

# Driver Code
if __name__ == "__main__":

    # Number of vertices
    n = 6

    # array of vectors is used
    # to store the graph
    # in the form of an adjacency list
    adj = [[] for _ in range(n)]

    # Given Graph
    add_edge(adj, 0, 1)
    add_edge(adj, 0, 2)
    add_edge(adj, 1, 3)
    add_edge(adj, 1, 4)
    add_edge(adj, 2, 3)
    add_edge(adj, 3, 5)
    add_edge(adj, 4, 5)

    # Given source and destination
    src = 0
    dest = n - 1

    # Function Call
    print_paths(adj, n, src, dest)

# This code is contributed by sanjeev2552

C#

/*package whatever //do not write package name here */
using System;
using System.Collections.Generic;

public class GFG
{

  // Function to form edge between
  // two vertices src and dest
  static void add_edge(List<List<int>> adj, int src, int dest){
    adj[src].Add(dest);
    adj[dest].Add(src);
  }

  // Function which finds all the paths
  // and stores it in paths array
  static void find_paths(List<List<int>> paths, List<int> path,
                         List<List<int>> parent, int n, int u) {
    // Base Case
    if (u == -1) {
      paths.Add(new List<int>(path));
      return;
    }

    // Loop for all the parents
    // of the given vertex
    foreach (int par in parent[u]) {

      // Insert the current
      // vertex in path
      path.Add(u);

      // Recursive call for its parent
      find_paths(paths, path, parent, n, par);

      // Remove the current vertex
      path.RemoveAt(path.Count-1);
    }
  }

  // Function which performs bfs
  // from the given source vertex
  static void bfs(List<List<int>> adj, List<List<int>> parent,
                  int n, int start) {

    // dist will contain shortest distance
    // from start to every other vertex
    int[] dist = new int[n];
    for(int i=0;i<n;i++)
      dist[i] = int.MaxValue;

    Queue<int> q = new Queue<int>();

    // Insert source vertex in queue and make
    // its parent -1 and distance 0
    q.Enqueue(start);

    parent[start].Clear();
    parent[start].Add(-1);

    dist[start] = 0;

    // Until Queue is empty
    while (q.Count!=0) {
      int u = q.Dequeue();

      foreach (int v in adj[u]) {
        if (dist[v] > dist[u] + 1) {

          // A shorter distance is found
          // So erase all the previous parents
          // and insert new parent u in parent[v]
          dist[v] = dist[u] + 1;
          q.Enqueue(v);
          parent[v].Clear();
          parent[v].Add(u);
        }
        else if (dist[v] == dist[u] + 1) {

          // Another candidate parent for
          // shortes path found
          parent[v].Add(u);
        }
      }
    }
  }

  // Function which prints all the paths
  // from start to end
  static void print_paths(List<List<int>> adj, int n, int start, int end){
    List<List<int>> paths = new List<List<int>>();
    List<int> path = new List<int>();
    List<List<int>> parent = new List<List<int>>();

    for(int i = 0; i < n; i++){
      parent.Add(new List<int>());
    }

    // Function call to bfs
    bfs(adj, parent, n, start);

    // Function call to find_paths
    find_paths(paths, path, parent, n, end);

    foreach (List<int> v in paths) {

      // Since paths contain each
      // path in reverse order,
      // so reverse it
      v.Reverse();

      // Print node for the current path
      foreach (int u in v)
        Console.Write(u + " ");

      Console.WriteLine();
    }
  }

  public static void Main(String[] args)
  {

    // Number of vertices
    int n = 6;

    // array of vectors is used
    // to store the graph
    // in the form of an adjacency list
    List<List<int>> adj = new List<List<int>>();
    for(int i = 0; i < n; i++){
      adj.Add(new List<int>());
    }

    // Given Graph
    add_edge(adj, 0, 1);
    add_edge(adj, 0, 2);
    add_edge(adj, 1, 3);
    add_edge(adj, 1, 4);
    add_edge(adj, 2, 3);
    add_edge(adj, 3, 5);
    add_edge(adj, 4, 5);

    // Given source and destination
    int src = 0;
    int dest = n - 1;

    // Function Call
    print_paths(adj, n, src, dest);
  }
}

// This code is contributed by shikhasingrajput

Javascript

// JavaScript program for the above approach

class Graph {
    // Function to form edge between two vertices src and dest
    addEdge(adj, src, dest) {
        adj[src].push(dest);
        adj[dest].push(src);
    }

    // Function which finds all the paths and stores it in paths array
    findPaths(paths, path, parent, n, u) {
        // Base Case
        if (u === -1) {
            paths.push(path.slice());
            return;
        }

        // Loop for all the parents of the given vertex
        for (let i = 0; i < parent[u].length; i++) {
            let par = parent[u][i];

            // Insert the current vertex in path
            path.push(u);

            // Recursive call for its parent
            this.findPaths(paths, path, parent, n, par);

            // Remove the current vertex
            path.pop();
        }
    }

    // Function which performs bfs from the given source vertex
    bfs(adj, parent, n, start) {
        // dist will contain shortest distance from start to every other vertex
        let dist = Array(n).fill(Number.MAX_VALUE);

        let q = [];

        // Insert source vertex in queue and make its parent -1 and distance 0
        q.push(start);
        parent[start] = [-1];
        dist[start] = 0;

        // Until Queue is empty
        while (q.length > 0) {
            let u = q.shift();

            for (let i = 0; i < adj[u].length; i++) {
                let v = adj[u][i];

                if (dist[v] > dist[u] + 1) {
                    // A shorter distance is found
                    // So erase all the previous parents
                    // and insert new parent u in parent[v]
                    dist[v] = dist[u] + 1;
                    q.push(v);
                    parent[v] = [u];
                } else if (dist[v] === dist[u] + 1) {
                    // Another candidate parent for shortes path found
                    parent[v].push(u);
                }
            }
        }
    }

    // Function which prints all the paths from start to end
    printPaths(adj, n, start, end) {
        let paths = [];
        let path = [];
        let parent = Array(n).fill(null).map(() => []);

        // Function call to bfs
        this.bfs(adj, parent, n, start);

        // Function call to findPaths
        this.findPaths(paths, path, parent, n, end);

        for (let i = 0; i < paths.length; i++) {
            let v = paths[i];

            // Since paths contain each path in reverse order, so reverse it
            v.reverse();

            // Print node for the current path
            console.log(v.join(" "));
        }
    }
}

let graph = new Graph();

// Number of vertices
let n = 6;

// Array to store the graph in the form of an adjacency list
let adj = [];
for (let i = 0; i < n; i++) {
    adj.push([]);
}

// Given graph
graph.addEdge(adj, 0, 1);
graph.addEdge(adj, 0, 2);
graph.addEdge(adj, 1, 3);
graph.addEdge(adj, 1, 4);
graph.addEdge(adj, 2, 3);
graph.addEdge(adj, 3, 5);
graph.addEdge(adj, 4, 5);

// Given source and destination
let src = 0;
let dest = n - 1;

// Function call
graph.printPaths(adj, n, src, dest);

// This code is contributed by lokeshmvs21.

Output:

0 1 3 5 
0 2 3 5 
0 1 4 5

Time Complexity: O(V + E) where V is the number of vertices and E is the number of edges. 
Auxiliary Space: O(V) where V is the number of vertices.