Question
Given the root of an n-ary tree, return the postorder 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)
Example 1:

Input: root = [1,null,3,2,4,null,5,6] Output: [5,6,3,2,4,1]
Example 2:

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: [2,6,14,11,7,3,12,8,4,13,9,10,5,1]
Constraints:
- The number of nodes in the tree is in the range
[0, 104]. 0 <= Node.val <= 104- The height of the n-ary tree is less than or equal to
1000.
Follow up: Recursive solution is trivial, could you do it iteratively?
Python Solution
"""
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""
class Solution:
def postorder(self, root: 'Node') -> List[int]:
if not root: return []
if not root.children: return [root.val]
l = []
for c in root.children: l += self.postorder(c)
return l + [root.val]

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