<p>You have <code>n</code> binary tree nodes numbered from <code>0</code> to <code>n - 1</code> where node <code>i</code> has two children <code>leftChild[i]</code> and <code>rightChild[i]</code>, return <code>true</code> if and only if <strong>all</strong> the given nodes form <strong>exactly one</strong> valid binary tree.</p>

<p>If node <code>i</code> has no left child then <code>leftChild[i]</code> will equal <code>-1</code>, similarly for the right child.</p>

<p>Note that the nodes have no values and that we only use the node numbers in this problem.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/08/23/1503_ex1.png" style="width: 195px; height: 287px;" />
<pre>
<strong>Input:</strong> n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/08/23/1503_ex2.png" style="width: 183px; height: 272px;" />
<pre>
<strong>Input:</strong> n = 4, leftChild = [1,-1,3,-1], rightChild = [2,3,-1,-1]
<strong>Output:</strong> false
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/08/23/1503_ex3.png" style="width: 82px; height: 174px;" />
<pre>
<strong>Input:</strong> n = 2, leftChild = [1,0], rightChild = [-1,-1]
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == leftChild.length == rightChild.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>-1 &lt;= leftChild[i], rightChild[i] &lt;= n - 1</code></li>
</ul>

class Solution {
public:
int find(int  x, vector<int>&parent){
      if(parent[x]==x)return x;

      return parent[x]=find(parent[x],parent);
}

void unionn(vector<int>&parent,vector<int>&rank,int x,int y){
    int x_parent=find(x,parent);
    int y_parent=find(y,parent);

    if(x_parent!=y_parent){
        if(rank[x_parent]>rank[y_parent]){
            parent[y_parent]=x_parent;
        }
         else if(rank[x_parent]<rank[y_parent]){
            parent[x_parent ]=y_parent;
        }
        else {
            parent[x_parent ]=y_parent;
            rank[y_parent]++;
        }
    }
}
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        unordered_map<int,list<int>>adjlist;
        vector<int>indegree(n,0);
        for(int i=0;i<n;i++){
            if(leftChild[i]!=-1){
                adjlist[i].push_back(leftChild[i]);
                indegree[leftChild[i]]++;
                }
            if(rightChild[i]!=-1){
            adjlist[i].push_back(rightChild[i]);
             indegree[rightChild[i]]++;
            }
        }
for(auto v: indegree){
    if(v>1)return false;
}
        vector<int>parent(n);
        for(int i=0;i<n;i++)parent[i]=i;
        vector<int>rank(n,0);

        for(int i=0;i<n;i++){
            
            for(auto nbr: adjlist[i]){
                int x=find(i,parent);
                int y=find(nbr,parent);
                if(x==y)return false;
                unionn(parent,rank,i,nbr);
            }
        }
        int first=find(0,parent);
        int count=0;
        for(int i=1;i<n;i++){
            if(find(i,parent)!=first)return false;
        }
        return true;
    }
};
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