590. N-ary Tree Postorder Traversal - Explanation

Problem Link

Description

You are 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:

  • 0 <= number of nodes in the tree <= 10,000
  • 0 <= Node.val <= 10,000
  • 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?


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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]:
        res = []

        def dfs(node):
            if not node:
                return

            for child in node.children:
                dfs(child)
            res.append(node.val)

        dfs(root)
        return res

Time & Space Complexity

  • Time complexity: O(n)O(n)
  • Space complexity: O(n)O(n) for the recursion stack.

2. Iterative DFS

"""
# 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 = []
        if not root:
            return res
        
        stack = [(root, False)]
        while stack:
            node, visited = stack.pop()
            if visited:
                res.append(node.val)
            else:
                stack.append((node, True))
                for child in reversed(node.children):
                    stack.append((child, False))
        return res

Time & Space Complexity

  • Time complexity: O(n)O(n)
  • Space complexity: O(n)O(n)