332. 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 {
    public List<String> findItinerary(List<List<String>> tickets) {
        if(tickets == null || tickets.size() == 0) return new ArrayList<>();
        List<String> res = new ArrayList<>();
        
        Map<String, PriorityQueue<String>> map = new HashMap<>();
        for(List<String> ticket : tickets){
            String from = ticket.get(0);
            String to = ticket.get(1);
            if(!map.containsKey(from)){
                map.put(from, new PriorityQueue<>());
            }
            map.get(from).offer(to);
        }
        
        dfs(map, res, "JFK");
        return res;
    }
    private void dfs(Map<String, PriorityQueue<String>> map, List<String> res, String curr){
        
        PriorityQueue<String> nexts = map.get(curr);
        while(nexts != null && !nexts.isEmpty()){
            String next = nexts.poll();
            dfs(map, res, next);
        }
        res.add(0, curr);
    }
}