0797. All Paths From Source to Target.py

# 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.


# 1) DFS
class Solution:
    def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:
        def dfs(u, path):
            if u == len(graph) - 1:
                paths.append(path)
                return
            for v in graph[u]:
                dfs(v, path + [v])

        paths = []
        dfs(0, [0])
        return paths


# 2) BFS
class Solution:
    def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:
        paths = []
        q = [[0]]
        while q:
            path = q.pop(0)
            u = path[-1]
            if u == len(graph) - 1:
                paths.append(path)
            for v in graph[u]:
                q.append(path + [v])
        return paths