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]