# Finding the path from one vertex to rest using BFS

Given an adjacency list representation of a directed graph, the task is to find the path from source to every other node in the graph using BFS.

**Examples:**

Input:Output:0 <- 2 1 <- 0 <- 2 2 3 <- 1 <- 0 <- 2 4 <- 5 <- 2 5 <- 2 6 <- 2

**Approach:** In the images shown below:

**que[]**array stores the vertices reached and we will**enqueue**a vertex only if it has not been visited and**dequeue**it once all its child node have been considered.- In order to distinguish whether the node has been visited or not we will put
**1**in**visited[]**array at the respective index to signify it has been visited and if at given index**0**is present it will signify that it has not been visited. - Parent array is to store the parent node of each vertex. For ex. In case of 0 connected to 2, 2 will be the parent node of 0 and we will put 2 at the index 0 in the parent array.

Below is the implementation of the above approach:

## Java

`// Java implementation of the approach ` `import` `java.util.ArrayList; ` `import` `java.util.Arrays; ` `import` `java.util.List; ` ` ` `class` `GFG { ` ` ` ` ` `// Function to print the path from ` ` ` `// source (s) to destination (d) ` ` ` `static` `void` `print(` `int` `parent[], ` `int` `s, ` `int` `d) ` ` ` `{ ` ` ` `// The while loop will stop only when the ` ` ` `// destination and the source node become equal ` ` ` `while` `(s != d) { ` ` ` ` ` `// Print the destination and store the parent ` ` ` `// of the node in the destination since parent ` ` ` `// stores the node through which ` ` ` `// the current node has been reached ` ` ` `System.out.print(d + ` `" <- "` `); ` ` ` `d = parent[d]; ` ` ` `} ` ` ` ` ` `System.out.println(d); ` ` ` `} ` ` ` ` ` `// Finding Path using BFS ALgorithm ` ` ` `static` `void` `bfs(List<List<Integer> > adjList, ` `int` `source, ` `int` `n) ` ` ` `{ ` ` ` `int` `parent[] = ` `new` `int` `[n]; ` ` ` `int` `que[] = ` `new` `int` `[n]; ` ` ` `Arrays.fill(parent, ` `0` `); ` ` ` `Arrays.fill(que, ` `0` `); ` ` ` ` ` `int` `front = -` `1` `, rear = -` `1` `; ` ` ` `int` `visited[] = ` `new` `int` `[n]; ` ` ` `Arrays.fill(visited, ` `0` `); ` ` ` `visited = ` `1` `; ` ` ` `parent = source; ` ` ` ` ` `// To add any non visited node we will increment the rear ` ` ` `// and add that vertex to the end of the array (enqueuing) ` ` ` `que[++rear] = source; ` ` ` ` ` `int` `k; ` ` ` ` ` `// The loop will continue till the rear and front are equal ` ` ` `while` `(front != rear) { ` ` ` ` ` `// Here Dequeuing is nothing but to increment the front int ` ` ` `k = que[++front]; ` ` ` `List<Integer> list = adjList.get(k); ` ` ` `for` `(` `int` `i = ` `0` `; i < list.size(); i++) { ` ` ` `int` `j = list.get(i); ` ` ` `if` `(visited[j] == ` `0` `) { ` ` ` `que[++rear] = j; ` ` ` `visited[j] = ` `1` `; ` ` ` `parent[j] = k; ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` ` ` `// Print the path from source to every other node ` ` ` `for` `(k = ` `0` `; k < n; k++) ` ` ` `print(parent, source, k); ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` ` ` `// Adjacency list representation of the graph ` ` ` `List<List<Integer> > adjList = ` `new` `ArrayList<>(); ` ` ` ` ` `// Vertices 1 and 2 have an incoming edge ` ` ` `// from vertex 0 ` ` ` `List<Integer> tmp = ` `new` `ArrayList<Integer>(Arrays.asList(` `1` `, ` `2` `)); ` ` ` `adjList.add(tmp); ` ` ` ` ` `// Vertex 3 has an incoming edge from vertex 1 ` ` ` `tmp = ` `new` `ArrayList<Integer>(Arrays.asList(` `3` `)); ` ` ` `adjList.add(tmp); ` ` ` ` ` `// Vertices 0, 5 and 6 have an incoming ` ` ` `// edge from vertex 2 ` ` ` `tmp = ` `new` `ArrayList<Integer>(Arrays.asList(` `0` `, ` `5` `, ` `6` `)); ` ` ` `adjList.add(tmp); ` ` ` ` ` `// Vertices 1 and 4 have an incoming edge ` ` ` `// from vertex 3 ` ` ` `tmp = ` `new` `ArrayList<Integer>(Arrays.asList(` `1` `, ` `4` `)); ` ` ` `adjList.add(tmp); ` ` ` ` ` `// Vertices 2 and 3 have an incoming edge ` ` ` `// from vertex 4 ` ` ` `tmp = ` `new` `ArrayList<Integer>(Arrays.asList(` `2` `, ` `3` `)); ` ` ` `adjList.add(tmp); ` ` ` ` ` `// Vertices 4 and 6 have an incoming edge ` ` ` `// from vertex 5 ` ` ` `tmp = ` `new` `ArrayList<Integer>(Arrays.asList(` `4` `, ` `6` `)); ` ` ` `adjList.add(tmp); ` ` ` ` ` `// Vertex 5 has an incoming edge from vertex 6 ` ` ` `tmp = ` `new` `ArrayList<Integer>(Arrays.asList(` `5` `)); ` ` ` `adjList.add(tmp); ` ` ` ` ` `int` `n = adjList.size(); ` ` ` ` ` `int` `source = ` `2` `; ` ` ` `bfs(adjList, source, n); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

## Python3

# Python3 implementation of the approach

# Function to print the path from

# src (s) to destination (d)

def printfunc(parent, s, d):

# The while loop will stop only when

# the destination and the src node

# become equal

while s != d:

# Print the destination and store

# the parent of the node in the

# destination since parent stores

# the node through which the current

# node has been reached

print(str(d) + ” <-", end = " ")
d = parent[d]
print(d)
# Finding Path using BFS ALgorithm
def bfs(adjList, src, n):
parent = [0] * (n)
que = [0] * (n)
front, rear = -1, -1
visited = [0] * (n)
visited[src] = 1
parent[src] = src
# To add any non visited node we will
# increment the rear and add that vertex
# to the end of the array (enqueuing)
rear += 1
que[rear] = src
# The loop will continue till the rear
# and front are equal
while front != rear:
# Here Dequeuing is nothing but to
# increment the front int
front += 1
k = que[front]
List = adjList[k]
for i in range(0, len(List)):
j = List[i]
if visited[j] == 0:
rear += 1
que[rear] = j
visited[j] = 1
parent[j] = k
# Print the path from src to every
# other node
for k in range(0, n):
printfunc(parent, src, k)
# Driver code
if __name__ == "__main__":
# Adjacency list representation
# of the graph
adjList = []
# Vertices 1 and 2 have an incoming edge
# from vertex 0
adjList.append([1, 2])
# Vertex 3 has an incoming edge
# from vertex 1
adjList.append([3])
# Vertices 0, 5 and 6 have an incoming
# edge from vertex 2
adjList.append([0, 5, 6])
# Vertices 1 and 4 have an incoming edge
# from vertex 3
adjList.append([1, 4])
# Vertices 2 and 3 have an incoming edge
# from vertex 4
adjList.append([2, 3])
# Vertices 4 and 6 have an incoming edge
# from vertex 5
adjList.append([4, 6])
# Vertex 5 has an incoming edge
# from vertex 6
adjList.append([5])
n = len(adjList)
src = 2
bfs(adjList, src, n)
# This code is contributed by Rituraj Jain
[tabbyending]

**Output:**

0 <- 2 1 <- 0 <- 2 2 3 <- 1 <- 0 <- 2 4 <- 5 <- 2 5 <- 2 6 <- 2

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