## Description

https://leetcode.com/problems/all-paths-from-source-to-target/

Given a directed acyclic graph (**DAG**) of `n`

nodes labeled from 0 to n – 1, find all possible paths from node `0`

to node `n - 1`

, and return them in any order.

The graph is given as follows: `graph[i]`

is a list of all nodes you can visit from node `i`

(i.e., there is a directed edge from node `i`

to node `graph[i][j]`

).

**Example 1:**

Input:graph = [[1,2],[3],[3],[]]Output:[[0,1,3],[0,2,3]]Explanation:There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.

**Example 2:**

Input:graph = [[4,3,1],[3,2,4],[3],[4],[]]Output:[[0,4],[0,3,4],[0,1,3,4],[0,1,2,3,4],[0,1,4]]

**Example 3:**

Input:graph = [[1],[]]Output:[[0,1]]

**Example 4:**

Input:graph = [[1,2,3],[2],[3],[]]Output:[[0,1,2,3],[0,2,3],[0,3]]

**Example 5:**

Input:graph = [[1,3],[2],[3],[]]Output:[[0,1,2,3],[0,3]]

**Constraints:**

`n == graph.length`

`2 <= n <= 15`

`0 <= graph[i][j] < n`

`graph[i][j] != i`

(i.e., there will be no self-loops).- The input graph is
**guaranteed**to be a**DAG**.

## Explanation

Build an adjacency list representing the relationship between nodes and edges. Then conduct a depth-first search to find all the paths from source to target.

## Python Solution

```
class Solution:
def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:
results = []
adjacency_list = []
for i in graph:
adjacency_list.append([])
for i in range(len(graph)):
edges = graph[i]
for edge in edges:
adjacency_list[i].append(edge)
self.helper(results, adjacency_list, 0, len(adjacency_list) - 1, [0])
return results
def helper(self, results, adjacency_list, current, target, path):
if current == target:
results.append(list(path))
return
for node in adjacency_list[current]:
path.append(node)
self.helper(results, adjacency_list, node, target, path)
path.pop()
```

- Time Complexity: O(V + E). V is the number of vertices, E is the number of edges.
- Space Complexity: O(V + E). V is the number of vertices, E is the number of edges.