human
Write me a the Python code you requested given the following:
I found an interesting problem on https://leetcode.com/problems/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.

Example 1:

Input: tickets = [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]

Example 2:

Input: tickets = [["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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python code you requested.

from collections import defaultdict
 
 class Solution:
     def findItinerary(self, tickets):
         """
         :type tickets: List[List[str]]
         :rtype: List[str]
         """
         
         graph = defaultdict(list)
         for from_, to_ in tickets:
             graph[from_].append(to_)
         
         for each in graph:
             graph[each].sort()
         
         res = []
         self.dfs(graph, "JFK", res)
         return res[::-1]
     
     
     def dfs(self, graph, from_, results):
         
         while graph[from_]:
             curr = graph[from_].pop(0)
             self.dfs(graph, curr, results)
         
         results.append(from_)