All Paths From Source to Target - Easy DFS solution

Backtracking     Breadth-First Search     C++     Depth-First Search     Graph     Medium    

Problem Statement:

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]]

 

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.

Solution:

 
class Solution {
public:
    void dfs(vector<vector<int>> graph, int u, int n, vector<int>&path,
            vector<vector<int>> &paths)
    {
        path.push_back(u);
        if (u==n-1) paths.push_back(path);
        for (int v: graph[u]) dfs(graph,v,n,path,paths);
        path.pop_back();
    }
    vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
        int n=graph.size();
        vector<int>path;
        vector<vector<int>> paths;
        dfs(graph,0,n,path,paths);
        return paths;
    }
};