<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) {
        map<int,int> visCntMap;
        set<int> nodeSet;
        for(auto& item:pairs){
            int a = item[0];
            int b = item[1];
            nodeSet.insert(a);
            nodeSet.insert(b);
            visCntMap[a]++;
            visCntMap[b]++;
        }
        map<int,vector<int>> tree;
        bool hasSame = false;
        for(auto& item:pairs){
            int a = item[0];
            int b = item[1];
            if(visCntMap[a] >= visCntMap[b]){
                //a is parent of b;
                tree[a].push_back(b);
                if(visCntMap[a] == visCntMap[b]){
                    hasSame = true;
                }
            }
            else{
                tree[b].push_back(a);
            }
        }
        map<int,int> parentCntMap;
        for(auto& item: tree){
            for(auto son: item.second){
                parentCntMap[son]++;
            }
        }
        int root = -1;
        int total = nodeSet.size();
        for(auto& item:visCntMap){
            if(item.second + 1 == total && parentCntMap.count(item.first) == 0){
                root = item.first;
                break;
            }
        }
        if(root == -1) return 0;
        queue<int> q;
        q.push(root);
        while(q.size() > 0){
            int node = q.front();
            q.pop();
            int visCnt = visCntMap[node];
            int cntFromSon = 0;
            for(auto son:tree[node]){
                visCntMap[son]--;
                parentCntMap[son]--;
                if(parentCntMap[son] == 0){
                    cntFromSon+= visCntMap[son] + 1;
                    q.push(son);
                }
            }
            if(visCnt != cntFromSon){
                return 0;
            }
        }
        for(auto& item:parentCntMap){
            if(item.second != 0) return 0;
        }
        return hasSame ? 2 : 1;


    }
};
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