<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(int src,vector<int> & parent,vector<int>& vis,vector<int> adj[])
    {
        queue<int> q;
        vis[src] = 1;
        parent[src] = -1;
        q.push(src);
        while(!q.empty())
        {
            int s = q.front(); q.pop();

            for(auto nbr : adj[s])
            {
                if(!vis[nbr])
                {
                    vis[nbr] = 1;
                    parent[nbr] = s;
                    q.push(nbr);
                }else if(nbr != parent[s])
                {
                    return true;
                }
            }
        }

        return false;
    }
    


    bool validateBinaryTreeNodes(int n, vector<int>& left, vector<int>& right) {
        
        vector<int> adj[n];
        map<int,int> vis;
        for(int i = 0 ;  i < n; i++)
        {
            if(left[i] != -1)
            {
                if(!vis[left[i]])
                {
                    vis[left[i]] = 1;
                    adj[i].push_back(left[i]);
                    adj[left[i]].push_back(i);
                }else return false;
            }

            if(right[i] != -1)
            {
                if(!vis[right[i]])
                {
                    vis[right[i]] = 1;
                    adj[i].push_back(right[i]);
                    adj[right[i]].push_back(i);
                }else return false;
            }
        }

        // check cycle
        vector<int> v(n,0),parent(n,0);
        int cnt = 0;
        for(int i = 0 ; i < n; i++)
        {
            if(v[i] == 0)
            {
                // cout<<"fff "<<i<<"\n";
                cnt++;
                if(cnt == 2) return false;
                bool f = bfs(i,parent,v,adj);
                // cout<<f<<"\n";
                if(f) return false;
            }
        }
        return true;
    }
};
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