797

797. All Paths From Source to Target (Medium)

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).
• All the elements of graph[i] are unique.
• The input graph is guaranteed to be a DAG.

Companies:
Bloomberg, Google, Microsoft, Apple, Amazon

Related Topics:
Backtracking, Depth-First Search, Breadth-First Search, Graph

Similar Questions:

• Number of Ways to Arrive at Destination (Medium)

Solution 1. DFS

Since it's an Acyclic Graph, we don't need to maintain a set of visited nodes to prevent going into a loop.

// OJ: https://leetcode.com/problems/all-paths-from-source-to-target/
// Author: github.com/lzl124631x
// Time: O(N * 2^N)
// Space: O(N) extra space
class Solution {
public:
    vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& G) {
        vector<int> path;
        vector<vector<int>> ans;
        int N = G.size();
        function<void(int)> dfs = [&](int u) {
            path.push_back(u);
            if (u == N - 1) ans.push_back(path);
            else {
                for (int v : G[u]) dfs(v);
            }
            path.pop_back();
        };
        dfs(0);
        return ans;
    }
};