LeetCode: 589. N-ary Tree Preorder Traversal

题目

原题地址:https://leetcode.com/problems/n-ary-tree-preorder-traversal/

Given an n-ary tree, return the preorder 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).

Follow up:

Recursive solution is trivial, could you do it iteratively?

Example 1:

image

Input: root = [1,null,3,2,4,null,5,6]
Output: [1,3,5,6,2,4]

Example 2:

image2

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

Constraints:

  • The height of the n-ary tree is less than or equal to 1000
  • The total number of nodes is between [0, 10^4]

解法

这个题目虽然说的是 N 叉树的前序遍历,但是实现方法跟二叉树的前序遍历几乎是一样的, 可以参考前面 LeetCode: 144. Binary Tree Preorder Traversal 的方法进行实现。

递归法实现

这个方法的 Python 代码类似下面这样:

"""
# Definition for a Node.
class Node(object):
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""

class Solution(object):
    def preorder(self, root):
        nodes = []
        self._preorder(root, nodes)

        return nodes

    def _preorder(self, root, nodes):
        if root is None:
            return

        nodes.append(root.val)

        for c in root.children:
            self._preorder(c, nodes)

stack 法实现

这个方法的 Python 代码类似下面这样:

"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""

class Solution:
    def preorder(self, root):
        if root is None:
            return []

        nodes = []
        stack = []
        stack.append(root)

        while stack:
            node = stack.pop()
            nodes.append(node.val)

            # 因为 stack 是后进先出,所以要反着来
            for c in node.children[::-1]:
                stack.append(c)

        return nodes

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