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

		// traverse for adjacent nodes
		for (auto it : adj[node]) {
			// when the node is not visited
			if (!vis[it]) {
				if (dfsCheck(it, adj, vis, pathVis) == true)
					return true;
			}
			// if the node has been previously visited
			// but it has to be visited on the same path
			else if (pathVis[it]) {
				return true;
			}
		}
		pathVis[node] = 0;
		return false;
	}
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        vector<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;
            }
            if(rightChild[i]!=-1){
              adj[i].push_back(rightChild[i]);
               indegree[rightChild[i]]++;
               if (indegree[rightChild[i]] > 1) return false;
            }
           
        }
      int root=-1;
       for(int i=0;i<n;i++){
        if(indegree[i]==0){
            if(root==-1){
                root=i;
            }
            else{
                return false;
            }
        }
       }
       if(root==-1) return false;
       vector<int> vis(n,0);
       vector<int> pathvis(n,0);
       if(dfsCheck(root, adj, vis, pathvis)) return false;
       for(auto i:vis){
        if(i==0) return false;
       }
       return true;
    }
};
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