<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 dfs(vector<int>adj[], map<int,int>&vis,int node)
    {
        vis[node]=1;
        for(auto e:adj[node])
        {
            if(vis.find(e)!=vis.end())
            {
                return false;
            }
            if(!dfs(adj,vis,e))
            return false;
        }
        return true;
    }
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        vector<int>adj[n];
        set<int>hasP;
        for(int i=0;i<n;i++)
        {
            if(leftChild[i]!=-1)
            {
                adj[i].push_back(leftChild[i]);
                hasP.insert(leftChild[i]);
            }
            if(rightChild[i]!=-1)
            {
                adj[i].push_back(rightChild[i]);
                hasP.insert(rightChild[i]);
            }
        }

        // for(int i=0;i<n;i++)
        // {
        //     for(auto e:adj[i])
        //     {
        //         cout<<e<<" ";
        //     }
        //     cout<<endl;
        // }

        int root=-1;
        for(int i=0;i<n;i++)
        {
            if(hasP.find(i)==hasP.end())
            {
                root=i;
                break;
            }
        }

        //cout<<root<<endl;

        if(root==-1)
        return false;

        map<int,int>vis;

        if(!dfs(adj,vis,root))
        return false;

        for(int i=0;i<n;i++)
        {
            if(vis.find(i)==vis.end())
            return false;
        }

        return true;
        



    }
};
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