<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:
void dfs(vector<vector<int>>& adj, vector<bool>& visited, int key, bool& check, int& node)
{
	visited[key] = true;
	++node;
	for (auto it : adj[key])
	{
		if (!visited[it])
			dfs(adj, visited, it, check, node);
		else
		{
			check = false;
			return;
		}
	}
}

bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild)
{
	vector<pair<int, int>> edges;
	for (int i = 0; i < n; ++i)
	{
		if (leftChild[i] != -1)
			edges.push_back({ i,leftChild[i] });
		if (rightChild[i] != -1)
			edges.push_back({ i,rightChild[i] });
	}
	vector<vector<int>> adj(n);
	vector<int> degree(n, 0);
	for (auto it : edges)
	{
		adj[it.first].push_back(it.second);
		degree[it.second]++;
	}
	int node = 0;
	bool check = true;
	vector<bool> visited(n, false);
	for (int i = 0; i < n; ++i)
	{
		if (degree[i] == 0)
		{
			dfs(adj, visited, i, check, node);
			if (node < n) return false;
			if (!check) return false;
			break;
		}
	}
	for (int i = 0; i < n; ++i)
		if (!visited[i]) return false;
	return true;
}
};
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