Reconstruct Itinerary

Given a list of airline tickets represented by pairs of departure and arrival airports [from, to], reconstruct the itinerary in order. All of the tickets belong to a man who departs from JFK. Thus, the itinerary must begin with JFK.

Note:

If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string. For example, the itinerary ["JFK", "LGA"] has a smaller lexical order than ["JFK", "LGB"].
All airports are represented by three capital letters (IATA code).
You may assume all tickets form at least one valid itinerary.
One must use all the tickets once and only once.
Example 1:

Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]
Example 2:

Input: [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"].
  But it is larger in lexical order.

class Solution {
public:
    unordered_map<string,priority_queue<string,vector<string>,std::greater<string>>> vec;

    vector<string> stk;

    void dfs(const string& curr)
    {
        while(vec.count(curr) && vec[curr].size() >0){
            string tmp =vec[curr].top();
            vec[curr].pop();
            dfs(move(tmp));//使用move函数，提高效率
        }
        //为了避免死胡同，一个节点的所有相邻节点访问完再入栈；
        //然后再逆序
        stk.emplace_back(curr);
    }
    vector<string> findItinerary(vector<vector<string>>& tickets) {
        for(auto& it:tickets){
            vec[it[0]].emplace(it[1]);
        }
        dfs("JFK");
        reverse(stk.begin(),stk.end());
        return stk;
    }
};

注意：通过逆序查找然后反向来避免进入死胡同。