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
class Solution:
    def validateBinaryTreeNodes(self, n: int, left_children: List[int], right_children: List[int]) -> bool:
        
        root = self.find_root(n, left_children, right_children)
        if root == -1:
            return False
        
        return self.helper(root, n, left_children, right_children)
        
    def find_root(self, n, left_children, right_children):
        left_counter = collections.Counter(left_children)
        right_counter = collections.Counter(right_children)
        for i in range(n):
            if i not in left_counter and i not in right_counter:
                return i
            
        return -1
        
    def helper(self, root, n, left_children, right_children):
        visited = set()
        if not self.dfs(root, left_children, right_children, visited):
            return False
        
        #print(visited)
        return len(visited) == n
        
    def dfs(self, root, left_children, right_children, visited):
        if root == -1:
            return True
        
        if root in visited:
            return False
        
        visited.add(root)
        left = self.dfs(left_children[root], left_children, right_children, visited)
        right = self.dfs(right_children[root], left_children, right_children, visited)
        
        return left and right

class Solution:
    def validateBinaryTreeNodes(self, n: int, left_children: List[int], right_children: List[int]) -> bool:
        
        root = self.find_root(n, left_children, right_children)
        if root == -1:
            return False
        
        return self.helper(root, n, left_children, right_children)
        
    def find_root(self, n, left_children, right_children):
        left_counter = collections.Counter(left_children)
        right_counter = collections.Counter(right_children)
        for i in range(n):
            if i not in left_counter and i not in right_counter:
                return i
            
        return -1
        
    def helper(self, root, n, left_children, right_children):
        visited = set()
        if not self.dfs(root, left_children, right_children, visited):
            return False
        
        #print(visited)
        return len(visited) == n
        
    def dfs(self, root, left_children, right_children, visited):
        if root == -1:
            return True
        
        if root in visited:
            return False
        
        visited.add(root)
        left = self.dfs(left_children[root], left_children, right_children, visited)
        right = self.dfs(right_children[root], left_children, right_children, visited)
        
        return left and right
