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:

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:

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 {
    Map<String, PriorityQueue<String>> graph = new HashMap<>();
    List<String> route = new LinkedList();
    // Just choose greedily at each port
    public List<String> findItinerary(List<List<String>> tickets) {
        for (List<String> ticket : tickets)
            graph.computeIfAbsent(ticket.get(0), k -> new PriorityQueue()).add(ticket.get(1));
        visit("JFK");
        return route;
    }

    void visit(String airport) {
        while (graph.containsKey(airport) && !graph.get(airport).isEmpty())
            visit(graph.get(airport).poll());
        route.add(0, airport);
    }
}