<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) {
            unordered_map<int,unordered_set<int>> adj;
          for(int i=0;i<pairs.size();++i){
            int node1=pairs[i][0];
            int node2=pairs[i][1];
            adj[node1].insert(node2);
            adj[node2].insert(node1);

          }
      
      
     priority_queue< pair<int,int> > pq;
      for(auto it:adj){
        int degree=it.second.size();
        pq.push({degree,it.first});

        

      }
     
      int ans=1;

      unordered_map<int,bool> visited; 

      while(pq.size()){
        auto  [degree,node]=pq.top();
           pq.pop();
        int parent=-1;
        int parentdegree=INT_MAX;
        visited[node]=true;
        for(int it:adj[node]){
        int neigh=it;
        int neighdegree=adj[neigh].size();
        if(visited[neigh]&&neighdegree<parentdegree){
            parentdegree=neighdegree;
            parent=neigh;
        }
    
        }

        if(parent==-1){
           
            if(degree!=adj.size()-1) return 0;
            continue;
        }
        cout<<node;

     // checking if cycle exist;

      for(int neigh:adj[node]){
            
            if(neigh==parent) continue;
            
            if(!adj[parent].count(neigh)){ return 0; }
         
      }

      if(parentdegree==degree) ans=2;
    


      }



          
    

   return ans;

    }
};
211500