题目如下:

Given a directed, acyclic graph of N nodes.  Find all possible paths from node 0 to node N-1, and return them in any order.
The graph is given as follows: the nodes are 0, 1, ..., graph.length - 1. graph[i] is a list of all nodes j for which the edge (i, j) exists. Example:
Input: [[1,2], [3], [3], []]
Output: [[0,1,3],[0,2,3]]
Explanation: The graph looks like this:
0--->1
| |
v v
2--->3
There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.


Note:
The number of nodes in the graph will be in the range [2, 15].
You can print different paths in any order, but you should keep the order of nodes inside one path.

解题思路:题目明确了没有回路,而且nodes的范围是[2,15],这实在是大大降低了难度,一个深度遍历就能搞定,注意由于结果需要输出path,所以需要记录每次遍历的过程。

代码如下:

class Solution(object):
def copy(self,src,dest):
for i in src:
dest.append(i)
return dest
def allPathsSourceTarget(self,graph):
stack = []
stack.append([0])
step = []
res = []
while len(stack) > 0:
path = stack.pop(0)
if path[-1] == len(graph) - 1:
res.append(self.copy(path,[]))
for i in graph[path[-1]]:
tl = self.copy(path,[])
tl.append(i)
stack.insert(0,tl)
return res

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