Given an n-ary tree, return the level order traversal of its nodes' values. Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).
Input: root = [1,null,3,2,4,null,5,6] Output: [,[3,2,4],[5,6]]
Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14] Output: [,[2,3,4,5],[6,7,8,9,10],[11,12,13],]
The height of the n-ary tree is less than or equal to 1000 The total number of nodes is between [0, 104]
class Solution: def levelOrder(self, root: 'Node') -> List[List[int]]: if not root: return  res =  level = [root] while level: res.append([i.val for i in level]) nex = [i for node in level for i in node.children if i] level = nex return res