## Topological Sorting

## Medium Topological Sorting

25%AcceptedGiven an directed graph, a topological order of the graph nodes is defined as follow:

- For each directed edge
`A -> B`

in graph, A must before B in the order list.- The first node in the order can be any node in the graph with no nodes direct to it.
Find any topological order for the given graph.

Have you met this question in a real interview?

YesExampleFor graph as follow:

The topological order can be:

`[0, 1, 2, 3, 4, 5] [0, 2, 3, 1, 5, 4] ...`

NoteYou can assume that there is at least one topological order in the graph.

ChallengeCan you do it in both BFS and DFS?

/** * 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<DirectedGraphNode*> ans; queue<DirectedGraphNode*> q; unordered_map<DirectedGraphNode*, int> inDegree; for(auto it = graph.begin(); it!= graph.end(); it++){ if(inDegree.find(*it) == inDegree.end()){ inDegree[*it] = 0; } for(auto neighbor = (*it)->neighbors.begin(); neighbor != (*it)->neighbors.end(); neighbor++){ if(inDegree.find(*neighbor) == inDegree.end()){ inDegree[*neighbor] = 1; }else{ inDegree[*neighbor]++; } } } for(auto it = inDegree.begin(); it != inDegree.end(); it++){ if(it->second == 0){ q.push(it->first); } } while(!q.empty()){ DirectedGraphNode * node = q.front(); q.pop(); ans.push_back(node); for(auto neighbor = node->neighbors.begin(); neighbor != node->neighbors.end(); neighbor++){ inDegree[*neighbor]--; if(inDegree[*neighbor] == 0){ q.push(*neighbor); } } } return ans; } };