<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:
        parent = [-1] * n  # Track the parent of each node
        uf = list(range(n))  # Union-Find parent array
    
        def find(x):
            if uf[x] != x:
                uf[x] = find(uf[x])
            return uf[x]
    
        def union(x, y):
            rootX = find(x)
            rootY = find(y)
            if rootX == rootY:
                return False  # Cycle detected
            uf[rootY] = rootX
            return True
    
        # Process all the edges
        for i in range(n):
            for child in (leftChild[i], rightChild[i]):
                if child != -1:
                    if parent[child] != -1:  # If a node has two parents
                        return False
                    parent[child] = i  # Mark the current node as its parent
                
                    if not union(i, child):  # Union-Find to detect cycle
                        return False
    
        # There should be exactly one root (a node with no parent)
        root_count = sum(p == -1 for p in parent)
        if root_count != 1:
            return False
    
        # Check connectivity: ensure all nodes are in one connected component
        root = find(0)
        for i in range(n):
            if find(i) != root:
                return False
    
        return True        
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