<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 bfs(map<int,vector<int>>&adj,int root,int n)
    {
        queue<int> q;
        q.push(root);
        int count=0;
        while(!q.empty())
        {
            int u=q.front();
            q.pop();
            count++;
            for(auto v:adj[u])
            {
                q.push(v);
            }
        }
        if(count==n)
        return true;
        return false;
        
    }
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        map<int,vector<int>>adj;
        map<int,int> child;
        for(int i=0;i<n;i++)
        {
            int l=leftChild[i];
            int r=rightChild[i];
            if(l!=-1)
            {
                adj[i].push_back(l);
                if(child.find(l)!=child.end())  return false;
                child[l]=i;
            }
             if(r!=-1)
            {
                adj[i].push_back(r);
                if(child.find(r)!=child.end())  return false;
                child[r]=i;
            }
        }
        int root=-1;
       for (int i = 0; i < n; i++) {
            if (child.find(i) == child.end()) {
                if (root != -1) return false; // More than one root found
                root = i;
            }
        }
        if(root==-1)   return false;
        return bfs(adj,root,n);
    }
};
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