590. N-ary Tree Postorder Traversal

Problem Link

Description

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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?

Solution

---

Python3

"""
# Definition for a Node.
class Node:
    def __init__(self, val: Optional[int] = None, children: Optional[List['Node']] = None):
        self.val = val
        self.children = children
"""
 
class Solution:
    def postorder(self, root: 'Node') -> List[int]:
        res = []
 
        def go(node):
            if not node: return
 
            for child in node.children:
                go(child)
            
            res.append(node.val)
 
        go(root)
        return res

Python

"""
# Definition for a Node.
class Node(object):
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""
class Solution(object):
    def postorder(self, root):
        """
        :type root: Node
        :rtype: List[int]
        """
        
        temp = []
        if root == None: return temp
        def recursion(root,temp):
            for child in root.children:
                recursion(child,temp)
            
            temp.append(root.val)
            
        recursion(root,temp)
        
        return temp