<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:
  def checkWays(self, pairs: List[List[int]]) -> int:
    kMax = 501
    graph = collections.defaultdict(list)
    degrees = [0] * kMax
    connected = [[False] * kMax for _ in range(kMax)]

    for u, v in pairs:
      graph[u].append(v)
      graph[v].append(u)
      degrees[u] += 1
      degrees[v] += 1
      connected[u][v] = True
      connected[v][u] = True

    # For each node, sort its children by degrees in descending order.
    for _, children in graph.items():
      children.sort(key=lambda a: degrees[a], reverse=True)

    # Find the root with a degree that equals to n - 1.
    root = next((i for i, d in enumerate(degrees) if d == len(graph) - 1), -1)
    if root == -1:
      return 0

    hasMoreThanOneWay = False

    def dfs(u: int, ancestors: List[int], seen: List[bool]) -> bool:
      """
      Returns True if each node rooted at u is connected to all of its
      ancestors.
      """
      nonlocal hasMoreThanOneWay
      seen[u] = True
      for ancestor in ancestors:
        if not connected[u][ancestor]:
          return False
      ancestors.append(u)
      for v in graph[u]:
        if seen[v]:
          continue
        if degrees[v] == degrees[u]:
          hasMoreThanOneWay = True
        if not dfs(v, ancestors, seen):
          return False
      ancestors.pop()
      return True

    if not dfs(root, [], [False] * kMax):
      return 0
    return 2 if hasMoreThanOneWay else 1
        
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