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.
Constraints:
The number of nodes in the graph will be in the range [2, 15].
You can print different paths in any order, but you should keep the order of nodes inside one path.
Solution:
class Solution {
public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
List<List<Integer>> res = new ArrayList();
int start = 0, end = graph.length - 1;
List<Integer> path = new ArrayList();
path.add(0);
bt(graph, start, end, path, res);
return res;
}
private void bt(int[][] graph, int curr, int end, List<Integer> path, List<List<Integer>> res) {
if (curr == end) {
res.add(new ArrayList(path));
return;
}
int[] nodes = graph[curr];
for (int i = 0; i < nodes.length; i ++) {
path.add(nodes[i]);
bt(graph, nodes[i], end, path, res);
path.remove(path.size() - 1);
}
}
}