human
I found an interesting problem on https://leetcode.com/problems/validate-binary-tree-nodes/:
You have n binary tree nodes numbered from 0 to n - 1 where node i has two children leftChild[i] and rightChild[i], return true if and only if all the given nodes form exactly one valid binary tree.
If node i has no left child then leftChild[i] will equal -1, similarly for the right child.
Note that the nodes have no values and that we only use the node numbers in this problem.
 
Example 1:

Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]
Output: true

Example 2:

Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,3,-1,-1]
Output: false

Example 3:

Input: n = 2, leftChild = [1,0], rightChild = [-1,-1]
Output: false

Example 4:

Input: n = 6, leftChild = [1,-1,-1,4,-1,-1], rightChild = [2,-1,-1,5,-1,-1]
Output: false

 
Constraints:

1 <= n <= 10^4
leftChild.length == rightChild.length == n
-1 <= leftChild[i], rightChild[i] <= n - 1

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python code you requested.

class Solution:
    def validateBinaryTreeNodes(self, n: int, leftChild: List[int], rightChild: List[int]) -> bool:
        indegree = [0]*n
        outdegree = [0]*n
        visited = [False]*n
        
        possible_roots = []
        for i in range(n):
            if leftChild[i] != -1:
                outdegree[i] += 1
                indegree[leftChild[i]] += 1
            if rightChild[i] != -1:
                outdegree[i] += 1
                indegree[rightChild[i]] += 1
    
        
        for i in range(n):
            if indegree[i] == 0 and outdegree[i] <= 2:
                possible_roots.append(i)
                # check for more than one root node
                if len(possible_roots) > 1:
                    return False

            if outdegree[i] > 2:
                return False
            
        if not possible_roots:
            return False
    
        queue = deque()
        queue.append(possible_roots[0])
        while queue:
            node = queue.popleft()
            if visited[node]:
                return False
            
            visited[node] = True
            
            lc = leftChild[node]
            rc = rightChild[node]
            if lc != -1:
                queue.append(lc)
            if rc != -1:
                queue.append(rc)
        
        
        return all(visited)