<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:
        # start from the root
        # root candidates will have len(nodes) - 1 indegrees

        # build graph
        graph = defaultdict(set)
        for u, v in pairs:
            graph[u].add(v)
            graph[v].add(u)
        
        def helper(nodes):
            # nodes = graph.keys()
            d = defaultdict(list) # degrees: node1, node2..., number of connected nodes (direct & indirect)
            m = len(nodes) - 1
            for node in nodes:
                d[len(graph[node])].append(node)

            # if there is no such node with m degree, just return 0, cannot build rooted tree
            if len(d[m]) == 0: 
                return 0
            
            root = d[m][0] # start with the first cand root
            # firstly remove the root in graph
            for node in graph[root]:
                graph[node].remove(root)
            
            # perform dfs on each sub trees, get the ways number 
            # dfs to find for each subtree i, what are the nodes in each subtree
            comps = defaultdict(set)
            seen = set()
            i = 0
            def dfs(node, i):
                comps[i].add(node)
                seen.add(node)
                for neighbor in graph[node]:
                    if neighbor not in seen:
                        dfs(neighbor, i)
            
            for node in nodes:
                if node != root and node not in seen:
                    dfs(node, i)
                    i += 1
            
            cand_ways = [helper(comps[i]) for i in comps]
            if 0 in cand_ways: return 0
            if 2 in cand_ways: return 2
            if len(d[m]) >= 2: return 2

            return 1
        
        return helper(set(graph.keys()))
            

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