<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) {
     vector<vector<bool>> graph(510, vector<bool>(510));
        vector<vector<int>> adjacencyList(510);
      
        // Initialize the graph with the given pairs.
        for (auto& pair : pairs) {
            int node1 = pair[0], node2 = pair[1];
            graph[node1][node2] = graph[node2][node1] = true;
            adjacencyList[node1].push_back(node2);
            adjacencyList[node2].push_back(node1);
        }
      
        // Create and populate a list of nodes present in the graph.
        vector<int> nodesList;
        for (int node = 1; node <= 500; ++node) {
            if (adjacencyList[node].size()) {
                nodesList.push_back(node);
                graph[node][node] = true; // A node is always connected to itself.
            }
        }
      
        // Sort the nodes based on the degree (number of neighbors), in ascending order.
        sort(nodesList.begin(), nodesList.end(), [&](int nodeX, int nodeY) -> bool {
            return adjacencyList[nodeX].size() < adjacencyList[nodeY].size();
        });

        // Will be used to detect if there are two nodes with the same degree.
        bool hasEqualDegreeNodes = false;
        // Counter to check how many possible root nodes we have.
        int rootCount = 0;
      
        // Iterate through each node to check if the graph can form a tree.
        for (int i = 0; i < nodesList.size(); ++i) {
            int currentNode = nodesList[i];
            int nextNodeIndex = i + 1;
          
            // Find the next node in the sorted list that is connected to the current node.
            while (nextNodeIndex < nodesList.size() && !graph[currentNode][nodesList[nextNodeIndex]]) {
                ++nextNodeIndex;
            }
          
            // If a connected node is found, validate further.
            if (nextNodeIndex < nodesList.size()) {
                int nextNode = nodesList[nextNodeIndex];
                // Check if the current and next nodes have the same degree.
                if (adjacencyList[currentNode].size() == adjacencyList[nextNode].size()) {
                    hasEqualDegreeNodes = true;
                }
              
                // Make sure all neighbors of the currentNode are connected to nextNode.
                for (int neighbor : adjacencyList[currentNode])
                    if (!graph[nextNode][neighbor])
                        return 0; // Return 0 if the neighbor is not connected to the next node.
              
            } else {
                // If no connected node is found, increment root count as it might be a root.
                ++rootCount;
            }
        }

        // If there is more than one root, it can’t form a tree.
        if (rootCount > 1) return 0;

        // If there are nodes with equal degrees, there are two ways to arrange the tree.
        if (hasEqualDegreeNodes) return 2;

        // If the execution reaches here, there is exactly one way to arrange the tree.
        return 1;
    }
};
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