# 332 Reconstruct Itinerary

## 1. Question

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. 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. 2.
All airports are represented by three capital letters (IATA code).
3. 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.

## 2. Implementation

(1) DFS
class Solution {
public List<String> findItinerary(String[][] tickets) {
List<String> route = new ArrayList<>();
if (tickets == null || tickets.length == 0) {
return route;
}
Map<String, PriorityQueue<String>> map = new HashMap<>();
for (String[] ticket : tickets) {
map.putIfAbsent(ticket[0], new PriorityQueue<>());
}
dfs("JFK", map, route);
return route;
}
public void dfs(String airport, Map<String, PriorityQueue<String>> map, List<String> route) {
while (map.containsKey(airport) && !map.get(airport).isEmpty()) {
String nextAirport = map.get(airport).remove();
dfs(nextAirport, map, route);
}