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+# [210. Course Schedule II](https://leetcode.com/problems/course-schedule-ii/)
+# 思路
+这题是`207. Course Schedule`的变体，207题要求判断是否可以完成整个课程，即判断有向图是否有环，然后我们用拓扑排序来判断是否有环。
+而这题其实就是求有向图的拓扑排序，也分为BFS和DFS两种，思路和207一样的，所以这里不再赘述，可参考[207题解](https://github.com/ShusenTang/LeetCode/blob/master/solutions/207.%20Course%20Schedule.md)。
+
+# C++
+## BFS
+``` C++
+class Solution {
+public:
+    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
+        int arc_num = prerequisites.size();
+        stack<int>stk;
+        vector<int>in_degree(numCourses, 0);
+        vector<vector<int>>G(numCourses, vector<int>{});
+        vector<int>res;
+        
+        for(auto &arc : prerequisites){ // 建图
+            in_degree[arc[0]]++;
+            G[arc[1]].push_back(arc[0]);
+        }
+        
+        for(int i = 0; i < numCourses; i++) // 先将所有入度为0的顶点入栈
+            if(!in_degree[i]) stk.push(i);
+        
+        while(!stk.empty()){
+            int course = stk.top(); stk.pop();
+            res.push_back(course);
+            for(int c: G[course]){ // 所有以course为起点的边的终点
+                if(!(--in_degree[c])) stk.push(c);
+            }
+        }
+        return res.size() == numCourses ? res : vector<int>{};
+    }
+};
+```
+
+## DFS
+``` C++
+class Solution {
+private:
+   bool DFS(vector<vector<int>>&G, vector<int>& visited, vector<int>& res, int i) {
+       if (visited[i] == -1) return false;
+       if (visited[i] == 1) return true;
+       visited[i] = -1;
+       for (auto a : G[i]) {
+           if (!DFS(G, visited, res, a)) return false;
+       }
+       res.push_back(i);
+       visited[i] = 1;
+       return true;
+   }
+public:
+   vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
+       vector<vector<int>> G(numCourses, vector<int>());
+       vector<int>visited(numCourses, 0);
+       vector<int>res;
+       
+       for (auto &arc : prerequisites) {
+           G[arc[1]].push_back(arc[0]);
+       }
+       
+       for (int i = 0; i < numCourses; ++i)
+           if (!DFS(G, visited, res, i)) return vector<int>{};
+       
+       // 注意最后要翻转一下
+       reverse(res.begin(), res.end());
+       return res;
+   }
+};
+```