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776. N Ary Tree Postorder Traversal

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Problem - N Ary Tree Postorder Traversal

Easy

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?

Solutions

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"""
# 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]:
        # result = []
        # def dfs(root):
        #     if root is None:
        #         return
        #     for child in root.children:
        #         dfs(child)
        #     result.append(root.val)
        #     return

        # dfs(root)
        # return result

        result = []
        if root is None:
            return result

        stack = [root]

        while stack:
            node = stack.pop()
            result.append(node.val)
            for child in node.children:
                stack.append(child)

        return result[::-1]

Submission Stats:

  • Runtime: 47 ms (71.00%)
  • Memory: 19.1 MB (99.24%)