How do you handle large N-ary trees in postorder traversal?

By using DFS and a stack, managing state efficiently, and ensuring the tree is processed iteratively to handle large inputs effectively.

What is the time complexity of the N-ary Tree Postorder Traversal?

The time complexity is O(m), where m is the number of nodes in the tree, as each node is processed once.

How do I implement N-ary Tree Postorder Traversal iteratively?

Use DFS with a stack, process the children in reverse order, and ensure that nodes are only visited after all their children have been processed.

What’s the best way to manage state during N-ary tree traversal?

Use a stack to manage the nodes and keep track of the traversal order. Make sure children are visited before their parent node to maintain postorder sequence.

What if the tree is empty? How should I handle that in the traversal?

If the tree is empty, return an empty list, as there are no nodes to traverse.

#590

Easy

auto_awesome二分·树·traversal

LeetCode 题解工作台

N 叉树的后序遍历

给定一个 n 叉树的根节点 root ，返回其节点值的后序遍历。 n 叉树在输入中按层序遍历进行序列化表示，每组子节点由空值 null 分隔（请参见示例）。示例 1：输入： root = [1,null,3,2,4,null,5,6] 输出： [5,6,3,2,4,1] 示例 2：输入…

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题目描述

给定一个 n 叉树的根节点 root ，返回其节点值的 后序遍历 。

n 叉树在输入中按层序遍历进行序列化表示，每组子节点由空值 null 分隔（请参见示例）。

示例 1：

输入：root = [1,null,3,2,4,null,5,6]
输出：[5,6,3,2,4,1]

示例 2：

输入：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]
输出：[2,6,14,11,7,3,12,8,4,13,9,10,5,1]

提示：

节点总数在范围 [0, 10⁴] 内
0 <= Node.val <= 10⁴
n 叉树的高度小于或等于 1000

进阶：递归法很简单，你可以使用迭代法完成此题吗?

lightbulb

解题思路

方法一：递归

我们可以递归地遍历整棵树。对于每个节点，先对该节点的每个子节点递归地调用函数，然后将节点的值加入答案。

时间复杂度 $O(n)$ ，空间复杂度 $O(n)$ 。其中 $n$ 为节点数。

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

        ans = []
        dfs(root)
        return ans

speed

复杂度分析

指标	值
时间	O(m)
空间	O(m)

psychology

面试官常问的追问

外企场景

question_mark
Ability to manage state during tree traversal
question_mark
Skill in implementing DFS using stack
question_mark
Efficiency in handling large trees with many nodes

warning

常见陷阱

外企场景

error
Not managing state correctly, leading to incorrect traversal order
error
Pushing children onto the stack in the wrong order
error
Missing edge case handling for an empty tree

swap_horiz

进阶变体

外企场景

arrow_right_alt
Iterative solution using DFS with explicit stack management
arrow_right_alt
Recursion-based solution for postorder traversal
arrow_right_alt
Optimized DFS for large N-ary trees with more than 1000 nodes

help

常见问题

外企场景

继续练习

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