590.n-ary-tree-postorder-traversal.py

#
# @lc app=leetcode id=590 lang=python3
#
# [590] N-ary Tree Postorder Traversal
#
# https://leetcode.com/problems/n-ary-tree-postorder-traversal/description/
#
# algorithms
# Easy (77.33%)
# Likes:    2091
# Dislikes: 91
# Total Accepted:    220.3K
# Total Submissions: 284.6K
# Testcase Example:  '[1,null,3,2,4,null,5,6]'
#
# 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:
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# 
# The number of nodes in the tree is in the range [0, 10^4].
# 0 <= Node.val <= 10^4
# 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?
# 
#

# @lc code=start
"""
# 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 []
        stk = [root]
        res = []
        while stk:
            node = stk.pop()
            res.append(node.val)
            # for child in node.children:
            #     stk.append(child)
            stk.extend(node.children)
        return res[::-1]
        
# @lc code=end