Description
https://leetcode.com/problems/all-paths-from-source-to-target/
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).- The input graph is guaranteed to be a DAG.
Explanation
Build an adjacency list representing the relationship between nodes and edges. Then conduct a depth-first search to find all the paths from source to target.
Python Solution
class Solution:
def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:
results = []
adjacency_list = []
for i in graph:
adjacency_list.append([])
for i in range(len(graph)):
edges = graph[i]
for edge in edges:
adjacency_list[i].append(edge)
self.helper(results, adjacency_list, 0, len(adjacency_list) - 1, [0])
return results
def helper(self, results, adjacency_list, current, target, path):
if current == target:
results.append(list(path))
return
for node in adjacency_list[current]:
path.append(node)
self.helper(results, adjacency_list, node, target, path)
path.pop()
- Time Complexity: O(V + E). V is the number of vertices, E is the number of edges.
- Space Complexity: O(V + E). V is the number of vertices, E is the number of edges.