<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(int node, map<int, vector<int>>& adj, vector<int>& vst) {
        vst[node] = 1; // grey
        for (auto v : adj[node]) {
            if (vst[v] == 1)
                return 1;
            else if (vst[v] == 0 and dfs(v, adj, vst)) {
                // if white(unvisited) and reached by path to a grey then its a
                // cycle
                return 1;
            }
        }
        vst[node] = 2; // black
        return 0;
    }
    bool validateBinaryTreeNodes(int n, vector<int>& left, vector<int>& right) {
        map<int, vector<int>> adj;
        vector<int> out(n, 0), in(n, 0);
        for (int i = 0; i < n; i++) {
            if (left[i] >= 0) {
                adj[i].push_back(left[i]);
                out[i]++;
                in[left[i]]++;
            }
            if (right[i] >= 0) {
                adj[i].push_back(right[i]);
                out[i]++;
                in[right[i]]++;
            }
        }
        int cnt = 0;
        if (n==1 and in[0]==0 and out[0]==0)
         return true;
        for (int i = 0; i < n; i++) {
            if (in[i]==0 and out[i]==0)
             return false;
            if (in[i] > 1 or cnt > 1)
                return false;
            if (in[i] == 0 and out[i] > 0)
                cnt++;
        } // eliminated possibilty of multiple tree and one child multiple
          // parent...
        // cheak for cycle
        cout << "cheaking for cycle...";
        vector<int> vst(n, 0);
        // bool isCycle =false;
        for (int i = 0; i < n; i++)
            if (dfs(i, adj, vst)) {
                cout <<i<<"\n";
                return false;
            }
        return true;
    }
};
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