<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 dfs(int node,vector<int>&vis,vector<int>&path,vector<int>adj[])
{
    vis[node]=1;
    path[node]=1;

    for(auto it: adj[node])
    {
        if(!vis[it])
        {

            if(dfs(it,vis,path,adj)==true)
            return true;
        }
        else
        if(path[it])
        {
return true;
        }

    }
    path[node]=0;
    return false;
}




    bool validateBinaryTreeNodes(int n, vector<int>& leftchild, vector<int>& rightchild) {
        //form adj list
        //detect a cycle

        vector<int> adj[n];
        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( indegree[leftchild[i]]>1)
        return false;
        }
        }
        for(int i=0;i<n;i++)
        {
             if(rightchild[i]!=-1)
        { adj[i].push_back(rightchild[i]);
        indegree[rightchild[i]]++;
        if( indegree[rightchild[i]]>1)
        return false;
        }
        }
        int count_of_root_node=0;
        for(int i=0;i<n;i++)
        {
            if(indegree[i]==0)
            count_of_root_node++;
        }
        if(count_of_root_node>1)
        return false;

        //detect a cycle in directed graph
        vector<int>path(n,0);
        vector<int>vis(n,0);
        int count_of_comp=0;
        for(int i=0;i<n;i++)
        {
            if(!vis[i])
            {
                if(dfs(i,vis,path,adj)==true)
                return false;
            }
        }
        int vis_nodes=0;
        for(int i=0;i<n;i++)
        {
            if(vis[i]!=0)
            vis_nodes++;

        }
        if(vis_nodes!=n)
        return false;
       return true; 
    }
};
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