<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:
  int checkWays(vector<vector<int>>& pairs) {
    constexpr int kMax = 501;
    unordered_map<int, vector<int>> graph;
    vector<int> degrees(kMax);
    vector<vector<bool>> connected(kMax, vector<bool>(kMax));

    for (const vector<int>& pair : pairs) {
      const int u = pair[0];
      const int v = pair[1];
      graph[u].push_back(v);
      graph[v].push_back(u);
      ++degrees[u];
      ++degrees[v];
      connected[u][v] = true;
      connected[v][u] = true;
    }

    // For each node, sort its children by degrees in descending order.
    for (auto& [_, children] : graph)
      ranges::sort(children, ranges::greater{},
                   [&degrees](int child) { return degrees[child]; });

    const int root = getRoot(degrees, graph.size());
    if (root == -1)
      return 0;
    if (!dfs(graph, root, degrees, connected, {}, vector<bool>(kMax)))
      return 0;
    return hasMoreThanOneWay ? 2 : 1;
  }

 private:
  bool hasMoreThanOneWay = false;

  // Returns the root by finding the node with a degree that equals to n - 1.
  int getRoot(const vector<int>& degrees, int n) {
    for (int i = 1; i < degrees.size(); ++i)
      if (degrees[i] == n - 1)
        return i;
    return -1;
  }

  // Returns true if each node rooted at u is connected to all of its ancestors.
  bool dfs(const unordered_map<int, vector<int>>& graph, int u,
           vector<int>& degrees, vector<vector<bool>>& connected,
           vector<int>&& ancestors, vector<bool>&& seen) {
    seen[u] = true;

    for (const int ancestor : ancestors)
      if (!connected[u][ancestor])
        return false;

    ancestors.push_back(u);

    for (const int v : graph.at(u)) {
      if (seen[v])
        continue;
      // We can swap u with v, so there are more than one way.
      if (degrees[v] == degrees[u])
        hasMoreThanOneWay = true;
      if (!dfs(graph, v, degrees, connected, std::move(ancestors),
               std::move(seen)))
        return false;
    }

    ancestors.pop_back();
    return true;
  }
};
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