<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>&parent,vector<int>&visited,vector<int>adj[]){
        visited[node]=1;
        for(auto it:adj[node]){
           if(!visited[it]){
            //parent[it]=node;
              if(!dfs(it,parent,visited,adj))return false;
           }else return false;
        }
        return true;
    }
    bool validateBinaryTreeNodes(int n, vector<int>& leftchild, vector<int>& rightchild) {
         vector<int> adj[n];
         vector<int>parent(n,-1);
         for(int i=0;i<n;i++){
            if(leftchild[i]>-1){
                adj[i].push_back(leftchild[i]); 
                if(parent[leftchild[i]]!=-1)return false;
                parent[leftchild[i]]=i;  
            }
            if(rightchild[i]>-1){
                adj[i].push_back(rightchild[i]);
                if(parent[rightchild[i]]!=-1)return false;
                 parent[rightchild[i]]=i;
            }
         }
         vector<int>visited(n,0);
         int root=-1;
         for(int i=0;i<n;i++){
            if(parent[i]==-1){
                if(root!=-1)return false;
                root=i;
            }
         }
         if(root==-1)return false;
         if(!dfs(root,parent,visited,adj))return false;
         for(int i=0;i<n;i++){
            if(!visited[i]){
                return false;
            }
         }
         return true;
    }
};
38967