797. All Paths From Source to Target

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797. All Paths From Source to Target

Description

Difficulty: Medium

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

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:

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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:

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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.

Hints/Notes

  • Traverse the graph, add node to the path when entering and remove the node from the path when exiting
  • DAG, i.e. directed acyclic graph, so there’s no cycle

Solution

Language: C++

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class Solution {
public:
    vector<vector<int>> res;

    vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
        vector<int> path;
        traverse(graph, 0, path);
        return res;
    }

    void traverse(vector<vector<int>>& graph, int n, vector<int>& path) {
        path.push_back(n);
        
        if (n == graph.size() - 1) {
            res.push_back(path);
        }

        for (int num : graph[n]) {
            traverse(graph, num, path);
        }    

        path.pop_back();
    }
};