<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:
    int findroot(vector<int>& lc , vector<int>& rc , int n){
        int root = -1 ;
        int cnt = 0 ;
        unordered_set<int> ch ;
        ch.insert(lc.begin() , lc.end() ) ;
        ch.insert(rc.begin() , rc.end()) ;

        for(int i = 0 ; i < n ; i++){
            if(ch.find(i) == ch.end()){
                root = i ;
                cnt++;
            }
        }

        if(cnt > 1){
            return -1 ;
        }
        return root ;
    }

    bool validateBinaryTreeNodes(int n, vector<int>& lc, vector<int>& rc) {
        int root = findroot(lc , rc , n) ;

        // if(n ==1 ){
        //     return true ;
        // }
        if(root == -1){
            return false ;
        }

        unordered_set<int> vis ;
        queue<int> pq ;
        pq.push(root) ;
        vis.insert(root) ;
        while(!pq.empty()){
            int node = pq.front() ;
            pq.pop() ;

            int ch[] = {lc[node] , rc[node] } ;
            for(auto it : ch){
                if(it !=-1){
                    if(vis.find(it) != vis.end()){
                        return false ;
                    }

                    pq.push(it) ;
                    vis.insert(it); 
                }
            }
        }

        return vis.size() == n ;
    }
};
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