# 797 All Paths From Source to Target

## 797. [All Paths From Source to Target](https://leetcode.com/problems/all-paths-from-source-to-target/description/)

## 1. Question

Given a directed, acyclic graph of`N`nodes. Find all possible paths from node`0`to node`N-1`, and return them in any order.

The graph is given as follows: the nodes are 0, 1, ..., graph.length - 1. graph\[i] is a list of all nodes j for which the edge (i, j) exists.

```
Example:
Input: [[1,2], [3], [3], []] 

Output: [[0,1,3],[0,2,3]] 

Explanation:
The graph looks like this:
0--->1
|    |
v    v
2--->3

There are two paths:
0 -> 1 ->3 and 0 -> 2 ->3.
```

## 2. Implementation

**(1) Backtracking**

```java
class Solution {
    public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
        List<List<Integer>> res = new ArrayList<>();

        if (graph == null || graph.length == 0) {
            return res;
        }

        int m = graph.length, n = graph[0].length;

        List<Integer> path = new ArrayList<>();
        path.add(0);
        getPaths(0, graph, path, res);
        return res;
    }

    public void getPaths(int curNode, int[][] graph, List<Integer> path, List<List<Integer>> res) {
        if (curNode == graph.length - 1) {
            res.add(new ArrayList<>(path));
            return;
        }

        for (int nextNode : graph[curNode]) {
            path.add(nextNode);
            getPaths(nextNode, graph, path, res);
            path.remove(path.size() - 1);
        }
    }
}
```

## 3. Time & Space Complexity

Backtracking: 时间复杂度O(N \* 2^N ), 总共有2^N个可能的路径, 而每个路径path.remove()需要O(N)的时间, 空间复杂度O(N \* 2^N)，总共有2 ^ N的路径，每个路径的递归深度最多为N


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