<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:
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild)
     {
          vector<int>adj[n];
           vector<int>visit(n,0);
            vector<int>indegree(n,0);
           for(int i=0;i<n;i++)
           {
               if(leftChild[i]!=-1)
                {
                   adj[i].push_back(leftChild[i]);
                   indegree[leftChild[i]]++;
                   if(leftChild[leftChild[i]]==i)
                     return false;
                }
                if(rightChild[i]!=-1)
                {
                  adj[i].push_back(rightChild[i]);
                  indegree[rightChild[i]]++;
                   if(rightChild[rightChild[i]]==i)
                     return false;
                }
           }  int count=0;
           queue<int>q;
             for(int i=0;i<n;i++)
             {
                 if(indegree[i]==0)
                  {
                   q.push(i);
                   visit[i]=1;
                    break;
                  }
             }
                // if(count>1)
                //   return false;
                int temp=0;
            while(!q.empty())
            {
                int u=q.front();
                q.pop();
                temp++;
                  for(auto v:adj[u])
                  {
                        if(visit[v])
                           return false;
                        else
                        {
                            visit[v]=1;
                            q.push(v);
                        }
                  }
            }  if(temp!=n)
                   return false;
              return true;
               
         
            
    }
};
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