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?
class Solution:
    def validateBinaryTreeNodes(self, n, left, right):
        roots={*range(n)}
        for x in left+right:
            if x==-1:
                continue
            if x not in roots:
                return False
            roots.discard(x)
        if len(roots)!=1:
            return False
        k=0
        stk=[roots.pop()]
        while stk:
            node=stk.pop()
            k+=1
            l,r=left[node],right[node]
            if l!=-1:
                stk.append(l)
            if r!=-1:
                stk.append(r)
        return k==n