<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:
        # what ARE the rules of a binary tree:
        # max two children (no need to check this given the inputs)
        # each tree must have 1 (no more, no less) from 0 to any node i.
        #   this is because a tree is connected and each node must have exactly 1 parent besides
        #   the root. 2 way to check: run a traversal from 0, see if we visit all nodes 0 through
        #   n-1 exactly once.

        graph = defaultdict(set)
        children = set()
        for parent, child in enumerate(leftChild):
            if child == -1:
                continue

            parent_edges = graph[parent]
            parent_edges.add(child)
            graph[parent] = parent_edges
            children.add(child)

        for parent, child in enumerate(rightChild):
            if child == -1:
                continue

            parent_edges = graph[parent]
            parent_edges.add(child)
            graph[parent] = parent_edges
            children.add(child)

        # find root to see what node to start traversal from
        root = None

        for node in range(n):
            if node not in children:
                root = node
                break
        
        if root is None:
            return False

        # use bfs for traversal of graph, each node should be seen once (check length of seen set
        # at the end and check if it matches n, during traversal return False if a seen node is 
        # visited again)
        
        level = [root]
        seen = set()

        while len(level) > 0:
            next_level = []
            
            for current in level:
                if current in seen:
                    return False

                seen.add(current)

                for neighbor in graph[current]:
                    if neighbor in seen:
                        return False
                    else:
                        next_level.append(neighbor)

            level = next_level

        if len(seen) == n:
            return True
        else:
            return False
        
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