<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>

vector<string> mitoa;
unordered_map<string, int> matoi;
bool myCmp(int a, int b) { return mitoa[a] > mitoa[b]; }

class Solution {
public:
    vector<string> findItinerary(vector<vector<string>>& tickets) {
        mitoa.clear();
        matoi.clear();
        matoi["JFK"] = 0;
        mitoa.push_back("JFK");
        vector<vector<int>> g(1);
        for(const auto& ticket: tickets)
        {
            int u, v;
            if(matoi.find(ticket[0]) == matoi.end())
            {
                matoi[ticket[0]] = u = mitoa.size();
                mitoa.push_back(ticket[0]);
                g.push_back(vector<int>());
            }
            else
                u = matoi[ticket[0]];

            if(matoi.find(ticket[1]) == matoi.end())
            {
                matoi[ticket[1]] = v = mitoa.size();
                mitoa.push_back(ticket[1]);
                g.push_back(vector<int>());
            }
            else
                v = matoi[ticket[1]];

            g[u].push_back(v);

        }

        for(auto & adjList: g)
            sort(adjList.begin(), adjList.end(), myCmp);

        stack<int> st;
        stack<int> path;

        int start = 0;

        st.push(start);
        while(st.size())
        {
            int curr = st.top();
            if(g[curr].size())
            {
                st.push(g[curr].back());
                g[curr].pop_back();
            }
            else
            {
                path.push(curr);
                st.pop();
            }
        }

        vector<string> ans(tickets.size() + 1);
        int i = 0;
        while(path.size())
        {
            start = path.top();
            path.pop();
            ans[i++] = mitoa[start];
        }

        return ans;
    }
};
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