<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 {
    unordered_map<string, vector<string>>graph;
    unordered_set<string>nodes;
    unordered_map<string,int>indegree, outdegree;
public:
    vector<string> findItinerary(vector<vector<string>>& tickets) {
        int n = tickets.size();
        if(!n)
            return {};
        for(const auto&t:tickets){
            graph[t[0]].push_back(t[1]);
            nodes.insert(t[0]);
            nodes.insert(t[1]);
            outdegree[t[0]]++;
            indegree[t[1]]++;
        }
        string start_point = "JFK";
        if(!validate(start_point) || start_point != "JFK")
            return {};
        for(auto&[_, row]:graph){
            sort(row.begin(), row.end(), greater<string>());
        }
        vector<string>ret;
        dfs("JFK", ret);
        reverse(ret.begin(), ret.end());
        return ret;
    }
    
    
    void dfs(const string& cur, vector<string>&ret){
        while(graph[cur].size()){
            auto nex = graph[cur].back();
            graph[cur].pop_back();
            dfs(nex, ret);
        }
        ret.push_back(cur);
    }
    
    bool validate(string& start_point){
        int start = 0, end = 0;
        for(const auto&node:nodes){
            if(abs(indegree[node]-outdegree[node])>1)
                return false;
            if(indegree[node]>outdegree[node])
                end++;
            else if(indegree[node]<outdegree[node]){
                start_point = node;
                start++;
            }
        }
        return (!start&&!end)||(start == 1 && end == 1);
    }
};
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