<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:
    def validateBinaryTreeNodes(self, n: int, leftChild: List[int], rightChild: List[int]) -> bool:
        n = len(leftChild)
        
        # tree has 2 requirements
        # 1. every node has a single parent
        # 2. single root (only one node has NO parent)
        # 3. no cycle
        # 4. all nodes connected to each other (single component)
        
        parent = [-1] * n
        
        # checking condition (1. and 2.)
        for idx, (left, right) in enumerate(zip(leftChild, rightChild)):
            
            if left != -1:
                # FAILED: condition (1.)
                if parent[left] != -1: return False
                parent[left] = idx
                
            if right != -1:
                # FAILED: condition (1.)
                if parent[right] != -1: return False
                parent[right] = idx
        
        # FAILED condition (2.)
        if parent.count(-1) != 1: return False
            
        # checking condition (3. and 4.)
        vis = set()
        def dfs_has_cycle(u):
            if u in vis:
                return True
            else:
                vis.add(u)
            
            for kid in [leftChild[u], rightChild[u]]:
                if kid != -1:
                    if dfs_has_cycle(kid): return True
            
        # FAILED condition (3.) - found a cycle
        if dfs_has_cycle(parent.index(-1)): return False
        
        # FAILED condition (4.) - DFS did not visit all nodes!
        if len(vis) != n: return False
        
        # did not FAIL any condition, success ;)
        return True

"""
Tricky test case (cycle and not connected):
4
[1,-1,3,-1]
[2,-1,-1,-1]

"""
23210