<p>You are given an array <code>pairs</code>, where <code>pairs[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>, and:</p>

<ul>
	<li>There are no duplicates.</li>
	<li><code>x<sub>i</sub> &lt; y<sub>i</sub></code></li>
</ul>

<p>Let <code>ways</code> be the number of rooted trees that satisfy the following conditions:</p>

<ul>
	<li>The tree consists of nodes whose values appeared in <code>pairs</code>.</li>
	<li>A pair <code>[x<sub>i</sub>, y<sub>i</sub>]</code> exists in <code>pairs</code> <strong>if and only if</strong> <code>x<sub>i</sub></code> is an ancestor of <code>y<sub>i</sub></code> or <code>y<sub>i</sub></code> is an ancestor of <code>x<sub>i</sub></code>.</li>
	<li><strong>Note:</strong> the tree does not have to be a binary tree.</li>
</ul>

<p>Two ways are considered to be different if there is at least one node that has different parents in both ways.</p>

<p>Return:</p>

<ul>
	<li><code>0</code> if <code>ways == 0</code></li>
	<li><code>1</code> if <code>ways == 1</code></li>
	<li><code>2</code> if <code>ways &gt; 1</code></li>
</ul>

<p>A <strong>rooted tree</strong> is a tree that has a single root node, and all edges are oriented to be outgoing from the root.</p>

<p>An <strong>ancestor</strong> of a node is any node on the path from the root to that node (excluding the node itself). The root has no ancestors.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://assets.leetcode.com/uploads/2020/12/03/trees2.png" style="width: 208px; height: 221px;" />
<pre>
<strong>Input:</strong> pairs = [[1,2],[2,3]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is exactly one valid rooted tree, which is shown in the above figure.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/03/tree.png" style="width: 234px; height: 241px;" />
<pre>
<strong>Input:</strong> pairs = [[1,2],[2,3],[1,3]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are multiple valid rooted trees. Three of them are shown in the above figures.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> pairs = [[1,2],[2,3],[2,4],[1,5]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no valid rooted trees.</pre>

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

<ul>
	<li><code>1 &lt;= pairs.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= x<sub>i </sub>&lt; y<sub>i</sub> &lt;= 500</code></li>
	<li>The elements in <code>pairs</code> are unique.</li>
</ul>

class Solution {
public:
    unordered_map<int, unordered_set<int>> graph;
    void buildGraph(vector<vector<int>>& pairs) {
        for(vector<int> pair: pairs) {
            graph[pair[0]].insert(pair[1]);
            graph[pair[1]].insert(pair[0]);
        }
    }
    int solve(unordered_set<int> nodes) {
        if(nodes.size() == 0) {
            return 1;
        }
        int n = nodes.size();
        vector<int> possibleRoots;
        for(int node: nodes) {
            if(graph.at(node).size() == n-1) possibleRoots.push_back(node);
        }

        if(possibleRoots.size() == 0) return 0;
        int root = possibleRoots.at(0);
        for(int node: graph[root]) {
            graph[node].erase(root);
        }

        unordered_set<int> seen;
        unordered_map<int, unordered_set<int>> components;

        function<void(int, int)> dfs = [&seen, this, root, &dfs, &components] (int node, int mark) {
            components[mark].insert(node);
            seen.insert(node);
            for(int neighbor: graph[node]) {
                if(node != root && seen.find(neighbor) == seen.end()) {
                    dfs(neighbor, mark);
                }
            }
        };
        int mark = 0;
        for(int node: nodes) {
            if(seen.find(node) == seen.end() && node != root) {
                dfs(node, mark);
                mark++;
            }
        }
        if(components.size() == 0) return 1;
        
        int numWays[components.size()];
        for(int i = 0; i < components.size(); i++) {
            numWays[i] = solve(components[i]);
        }

        int ans = 1;
        for(int i = 0; i < components.size(); i++) {
            if(numWays[i] == 0) return 0;
            if(numWays[i] == 2) ans = 2;
        }

        if(ans == 2) return 2;
        if(possibleRoots.size() > 1) return 2;
        return 1;
    }
    int checkWays(vector<vector<int>>& pairs) {
        buildGraph(pairs);
        unordered_set<int> nodes;
        for(auto pair: graph) {
            nodes.insert(pair.first);
        }
        return solve(nodes);
    }


};
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