LeetCode: 590. N-ary Tree Postorder Traversal

题目

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

Given 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).

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: [5,6,3,2,4,1]

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

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 叉树的后序遍历,但是实现方法跟二叉树的后序遍历几乎是一样的, 可以参考前面 LeetCode: 145. Binary Tree Postorder 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 postorder(self, root):
        nodes = []
        self._postorder(root, nodes)
        return nodes

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

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

        nodes.append(root.val)

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 postorder(self, root):
        if root is None:
            return []

        nodes = []
        stack1 = []
        stack2 = []

        stack1.append(root)
        while stack1:
            curr = stack1.pop()
            stack2.append(curr)

            for c in curr.children:
               stack1.append(c)

        while stack2:
            node = stack2.pop()
            nodes.append(node.val)

        return nodes

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