<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:
 void dfs(int u, vector<int>adj[], vector<int>&vis){
    vis[u]=1;
    for(auto it:adj[u]){
        if(!vis[it]) dfs(it, adj, vis);
    }
 }

bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
    vector<int> adj[n];
    vector<int> in_degree(n, 0);
    
    for (int i = 0; i < n; i++) {
        if (leftChild[i] != -1) {
            adj[i].push_back(leftChild[i]);
            in_degree[leftChild[i]]++;
        }
        if (rightChild[i] != -1) {
            adj[i].push_back(rightChild[i]);
            in_degree[rightChild[i]]++;
        }
    }

    int node;
    vector<int>vis(n, 0);
    for(int i=0; i<n; i++){
       //cout<<in_degree[i]<<endl;
         if(in_degree[i]==0){
            node=i;
             dfs(i, adj, vis);
         }
    }
    for(int i=0; i<n; i++) if(!vis[i]) return false;
    
    for(int i=0; i<n ; i++){
        if(i!=node){
            if(in_degree[i]!=1) return false;
        }
    }
 return true;














    // vector<int>vis(n, 0);
    // vis[q.front()]=1;
    // while(!q.empty()){
    //     int t=q.front();
    //     q.pop();
        
    //     for(auto it:adj[t]){
    //         in_degree[it]--;
    //         if(in_degree[it]==0){ q.push(it); vis[it]=1;}

    //     }
    // }
    // return cnt==n;
}

};
44007