<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:
        # Step 1: Initialize a parent count array
        parent = [-1] * n  # parent[i] is -1 if i has no parent, otherwise stores the parent of i
        
        # Step 2: Check parent relationships and ensure no node has more than one parent
        for i in range(n):
            # Check left child
            if leftChild[i] != -1:
                if parent[leftChild[i]] != -1:
                    return False  # leftChild[i] already has a parent, invalid
                parent[leftChild[i]] = i  # Set i as the parent of leftChild[i]
            
            # Check right child
            if rightChild[i] != -1:
                if parent[rightChild[i]] != -1:
                    return False  # rightChild[i] already has a parent, invalid
                parent[rightChild[i]] = i  # Set i as the parent of rightChild[i]
        
        # Step 3: Find the root node (it should have no parent)
        root = -1
        for i in range(n):
            if parent[i] == -1:  # Node with no parent
                if root == -1:
                    root = i  # Found the root
                else:
                    return False  # More than one root found, invalid
        
        # If no root is found, return False
        if root == -1:
            return False
        
        # Step 4: Perform BFS or DFS to check if all nodes are connected to the root
        visited = set()
        
        def dfs(node):
            if node == -1 or node in visited:
                return
            visited.add(node)
            dfs(leftChild[node])
            dfs(rightChild[node])
        
        # Start DFS from the root
        dfs(root)
        
        # Step 5: Check if we visited all nodes
        return len(visited) == n

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