拓扑排序（BFS，DFS）

我们用两种方法来做这道题。

BFS

宽度优先搜索，是说我们每次把没有父亲节点的节点打印。具体操作可以是计算所有节点的入度。将入度为0的节点输出，同时将输出节点的儿子的入度减去一。我们用队列维持入度为0的节点集合。

/** * Definition for Directed graph. * struct DirectedGraphNode { *     int label; *     vector<DirectedGraphNode *> neighbors; *     DirectedGraphNode(int x) : label(x) {}; * }; */class Solution {public:    /**     * @param graph: A list of Directed graph node     * @return: Any topological order for the given graph.     */    vector<DirectedGraphNode*> topSort(vector<DirectedGraphNode*> graph) {        // write your code here        vector<int> indegree(graph.size(),0);        for(auto node:graph){            for(int i = 0; i < node->neighbors.size(); ++i){                indegree[node->neighbors[i]->label]++;            }        }        queue<DirectedGraphNode*> que;        for(int i = 0; i < indegree.size(); ++i){            if(indegree[i] == 0) que.push(graph[i]);        }        vector<DirectedGraphNode *> res;        while(!que.empty()){            DirectedGraphNode *temp = que.front();            for(auto pi:temp->neighbors){                indegree[pi->label]--;                if(indegree[pi->label] == 0) que.push(graph[pi->label]);            }            que.pop();            res.push_back(temp);        }        return res;    }};

DFS

深度优先搜索则是一旦我们要输出一个节点，则把它的父亲全部输出即可。考虑到我们的数据结构是记录每个节点的儿子节点，所以我们这样考虑：
一旦要输出一个节点，先把儿子节点全部输出；最后reverse即可。
创建childinres布尔数组是为了查询O(1)。《—–感谢roommate

/** * Definition for Directed graph. * struct DirectedGraphNode { *     int label; *     vector<DirectedGraphNode *> neighbors; *     DirectedGraphNode(int x) : label(x) {}; * }; */class Solution {public:    /**     * @param graph: A list of Directed graph node     * @return: Any topological order for the given graph.     */    void dfs(vector<DirectedGraphNode*> &res, vector<DirectedGraphNode*>& graph, int target, vector<bool>& targetinres){        if(targetinres[target]) return;        //cout<<target<<endl;        for(auto node:graph[target]->neighbors)            dfs(res,graph,node->label,targetinres);        res.push_back(graph[target]);        targetinres[target] = true;    }    vector<DirectedGraphNode*> topSort(vector<DirectedGraphNode*> graph) {        // write your code here        vector<DirectedGraphNode*> res;        vector<bool> targetinres(graph.size(),false);        for(int i = 0; i < graph.size(); ++i){            dfs(res,graph,i,targetinres);        }        reverse(res.begin(),res.end());        return res;    }};