<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>

from collections import defaultdict
from typing import List

class Solution:
    def checkWays(self, pairs: List[List[int]]) -> int:
        kMax = 501
        graph = defaultdict(list)
        degrees = [0] * kMax
        connected = [[False] * kMax for _ in range(kMax)]
        
        # Build the graph and degrees
        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
        
        # Sort neighbors by degree in descending order
        for u in graph:
            graph[u].sort(key=lambda x: -degrees[x])
        
        # Find the root node
        root = self.getRoot(degrees, len(graph))
        if root == -1:
            return 0
        
        # Depth-First Search (DFS) to check the conditions
        self.hasMoreThanOneWay = False
        if not self.dfs(graph, root, degrees, connected, [], [False] * kMax):
            return 0
        
        return 2 if self.hasMoreThanOneWay else 1
    
    def getRoot(self, degrees: List[int], n: int) -> int:
        for i in range(1, len(degrees)):
            if degrees[i] == n - 1:
                return i
        return -1
    
    def dfs(self, graph: defaultdict(int), u: int, degrees: List[int],
            connected: List[List[bool]], ancestors: List[int], seen: List[bool]) -> bool:
        seen[u] = True
        
        # Check if all ancestors are connected to u
        for ancestor in ancestors:
            if not connected[u][ancestor]:
                return False
        
        ancestors.append(u)
        
        for v in graph[u]:
            if seen[v]:
                continue
            # We can swap u with v, so there are more than one way.
            if degrees[v] == degrees[u]:
                self.hasMoreThanOneWay = True
            if not self.dfs(graph, v, degrees, connected, ancestors, seen):
                return False
        
        ancestors.pop()
        return True

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