Problem

Given the root of an n-ary tree, return the preorder 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: [1,3,5,6,2,4]

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: [1,2,3,6,7,11,14,4,8,12,5,9,13,10]

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?

Solution

"""
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""

class Solution:
def preorder(self, root: 'Node') -> List[int]:
# pre order means we travel all the way down the left before we go to the next node over
# so it's a DFS
self.output = []
def dfs(root):
# basecase
if root is None:
return
self.output.append(root.val)
for child in root.children:
dfs(child)
dfs(root)
return self.output