Topological Sorting

Question:
http://www.lintcode.com/en/problem/topological-sorting/
Given 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.

Answer:

/**
 * 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
        int len = graph.size();
        map<DirectedGraphNode*, int> indegree;
        for (auto node : graph)
            indegree[node] = 0;
        for (auto node : graph) {
            for (auto nn : node->neighbors) {
                indegree[nn] = indegree[nn] + 1;
            }
        }
        stack<DirectedGraphNode*> st;
        for (auto dp : indegree) {
            if (dp.second == 0)
                st.push(dp.first);
        }
        vector<DirectedGraphNode*> result;
        while (!st.empty()) {
            DirectedGraphNode* node = st.top();
            result.push_back(node);
            st.pop();
            for (auto nn : node->neighbors) {
                indegree[nn] = indegree[nn] - 1;
                if (indegree[nn] == 0)
                    st.push(nn);
            }
        }
        return result;
    }
};