<p>You are given a list of airline <code>tickets</code> where <code>tickets[i] = [from<sub>i</sub>, to<sub>i</sub>]</code> represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.</p>

<p>All of the tickets belong to a man who departs from <code>&quot;JFK&quot;</code>, thus, the itinerary must begin with <code>&quot;JFK&quot;</code>. If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string.</p>

<ul>
	<li>For example, the itinerary <code>[&quot;JFK&quot;, &quot;LGA&quot;]</code> has a smaller lexical order than <code>[&quot;JFK&quot;, &quot;LGB&quot;]</code>.</li>
</ul>

<p>You may assume all tickets form at least one valid itinerary. You must use all the tickets once and only once.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/itinerary1-graph.jpg" style="width: 382px; height: 222px;" />
<pre>
<strong>Input:</strong> tickets = [[&quot;MUC&quot;,&quot;LHR&quot;],[&quot;JFK&quot;,&quot;MUC&quot;],[&quot;SFO&quot;,&quot;SJC&quot;],[&quot;LHR&quot;,&quot;SFO&quot;]]
<strong>Output:</strong> [&quot;JFK&quot;,&quot;MUC&quot;,&quot;LHR&quot;,&quot;SFO&quot;,&quot;SJC&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/itinerary2-graph.jpg" style="width: 222px; height: 230px;" />
<pre>
<strong>Input:</strong> tickets = [[&quot;JFK&quot;,&quot;SFO&quot;],[&quot;JFK&quot;,&quot;ATL&quot;],[&quot;SFO&quot;,&quot;ATL&quot;],[&quot;ATL&quot;,&quot;JFK&quot;],[&quot;ATL&quot;,&quot;SFO&quot;]]
<strong>Output:</strong> [&quot;JFK&quot;,&quot;ATL&quot;,&quot;JFK&quot;,&quot;SFO&quot;,&quot;ATL&quot;,&quot;SFO&quot;]
<strong>Explanation:</strong> Another possible reconstruction is [&quot;JFK&quot;,&quot;SFO&quot;,&quot;ATL&quot;,&quot;JFK&quot;,&quot;ATL&quot;,&quot;SFO&quot;] but it is larger in lexical order.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= tickets.length &lt;= 300</code></li>
	<li><code>tickets[i].length == 2</code></li>
	<li><code>from<sub>i</sub>.length == 3</code></li>
	<li><code>to<sub>i</sub>.length == 3</code></li>
	<li><code>from<sub>i</sub></code> and <code>to<sub>i</sub></code> consist of uppercase English letters.</li>
	<li><code>from<sub>i</sub> != to<sub>i</sub></code></li>
</ul>

class Solution {
private:
    void eulerianPath(
        unordered_map<string, vector<string>>& graph,
        unordered_map<string, int>& outDegree,
        string& from,
        stack<string>& itinerary
    ) {
        while (outDegree[from] != 0) {
            string to = *graph[from].begin();
            graph[from].erase(graph[from].begin());
            outDegree[from]--;
            eulerianPath(graph, outDegree, to, itinerary);
        }
        itinerary.push(from);
    }

public:
    // Eulerian path solution: https://leetcode.com/problems/reconstruct-itinerary/submissions/1343666189?envType=study-plan-v2&envId=google-spring-23-high-frequency

    // See https://www.youtube.com/watch?v=8MpoO2zA2l4 for Eulerian Path
    vector<string> findItinerary(vector<vector<string>>& tickets) {
        unordered_map<string, vector<string> > graph;
        // unordered_map<string, int> inDegree;
        unordered_map<string, int> outDegree;

        for (int i = 0; i < tickets.size(); i++) {
            string from = tickets[i][0];
            string to = tickets[i][1];
            graph[from].push_back(to);
            // inDegree[to]++;
            outDegree[from]++;
        }

        for (auto& it: graph) {
            sort(it.second.begin(), it.second.end());
        }

        stack<string> reverseItinerary;
        string startNode = "JFK";
        eulerianPath(graph, outDegree, startNode, reverseItinerary);

        vector<string> itinerary;
        while (!reverseItinerary.empty()) {
            itinerary.push_back(reverseItinerary.top());
            reverseItinerary.pop();
        }
        return itinerary;
    }
};
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