[Leetcode] 797. All Paths From Source to Target 解题报告

题目：

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.
思路：

典型的DFS+Backtracking的题目：我们从0号节点开始进行DFS搜索，一旦发现达到了n-1号节点，就输出。当然在遍历到中间结点的时候，还需要回溯，以便于返回所有符合条件的路径。

代码：

class Solution {
public:
vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
vector<vector<int>> ret;
vector<int> line;
vector<bool> visited(graph.size(), false);
DFS(graph, visited, ret, line, 0);
return ret;
}
private:
void DFS(vector<vector<int>> &graph, vector<bool> &visited,
vector<vector<int>> &ret, vector<int> &line, int index) {
int n = graph.size();
if (index == n - 1) { // reaches to the target
ret.push_back(line);
ret.back().push_back(index);
return;
}
if (visited[index]) { // this node has already been visited
return;
}
line.push_back(index); // mark it as visited
visited[index] = true;
for (int i = 0; i < graph[index].size(); ++i) {
DFS(graph, visited, ret, line, graph[index][i]);
}
line.pop_back(); // backtracking
visited[index] = false;
}
};