<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 = list(range(n))
        rank = [0] * n
        
        def find(x):
            if parent[x] != x:
                parent[x] = find(parent[x])
            return parent[x]
        
        def union(x, y):
            px, py = find(x), find(y)
            if px == py:
                return False  # Cycle detected
            if rank[px] > rank[py]:
                parent[py] = px
            elif rank[px] < rank[py]:
                parent[px] = py
            else:
                parent[py] = px
                rank[px] += 1
            return True
        
        # Check for nodes with multiple parents
        has_parent = [False] * n
        for i in range(n):
            for child in [leftChild[i], rightChild[i]]:
                if child != -1:
                    if has_parent[child]:
                        return False  # Node already has a parent
                    has_parent[child] = True
                    if not union(i, child):
                        return False  # Cycle detected
        
        # Check if there is exactly one root
        root_count = 0
        for i in range(n):
            if parent[i] == i:
                root_count += 1
                if root_count > 1:
                    return False  # More than one root
        
        # Check if all nodes are connected
        root = find(0)
        for i in range(1, n):
            if find(i) != root:
                return False  # Not all nodes are connected
        
        return True

# Example usage:
solution = Solution()
n = 4
leftChild = [1, -1, 3, -1]
rightChild = [2, -1, -1, -1]
print(solution.validateBinaryTreeNodes(n, leftChild, rightChild))  # Output: True
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