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:

1. 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"].
2. All airports are represented by three capital letters (IATA code).
3. You may assume all tickets form at least one valid itinerary.

 

Example 1:

tickets = [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]

Return ["JFK", "MUC", "LHR", "SFO", "SJC"].

Example 2:

tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]

Return ["JFK","ATL","JFK","SFO","ATL","SFO"].

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

重建行程单，在图中找一条路径，能经过所有的边。

参考：

class Solution{public:    vector
     
       findItinerary(vector
      
       
        > tickets){\        unordered_map
        
         
          > m;        for(auto t : tickets){            m[t.first].insert(t.second);        }        vector
          
            res; dfs(m,"JFK",res); return vector
           
             (res.rbegin(),res.rend()); } void dfs(unordered_map
            
             
              >& m,string s,vector
              
               & res){ while(m[s].size()){ string t = *m[s].begin(); m[s].erase(m[s].begin()); dfs(m,t,res); } res.push_back(s); }};