<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>

struct Tree{
    int val,p;
    Tree* left;
    Tree* right;
    Tree(int x, int y, Tree* l, Tree* r): val(x), p(y), left(l), right(r) {}
    
};
class Solution {
public:
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        vector<Tree*>a;
        int i=0,c=0,r=-1,s=0;
        for(i=0;i<n;i++)
        {
            Tree* t=new Tree(-1,i,NULL,NULL);
            a.push_back(t);
        }
        for(i=0;i<n;i++)
        {
            if(leftChild[i]!=-1) 
            {
                a[i]->left=a[leftChild[i]];
                a[leftChild[i]]->val=i;
            }
            if(rightChild[i]!=-1) 
            {
                a[i]->right=a[rightChild[i]];
                a[rightChild[i]]->val=i;
            }
        }
        for(i=0;i<a.size();i++)
            if(a[i]->val==-1)
            {
                c++;
                r=i;
            }
        if(c!=1) return false;
        vector<int>b(n,-1);
        queue<Tree*>q;
        q.push(a[r]);
        while(q.size()>0)
        {
            s=q.size();
            for(i=0;i<s;i++)
            {
                Tree* t=q.front();
                if(b[t->p]!=-1) return false;
                b[t->p]=0;
                if(t->left!=NULL) q.push(t->left);
                if(t->right!=NULL) q.push(t->right);
                q.pop();
            }
        }
        for(i=0;i<b.size();i++)
            if(b[i]==-1) return false;
        return true;
    }
};
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