[leetcode] 590. N-ary Tree Postorder Traversal

Description

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:

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

分析

题目的意思是：求n叉树的后续遍历，后续遍历是左右中，所以在递归的时候等孩子节点遍历完以后，最后放入结果集合里面。迭代的方法需要用到栈，孩子节点从左到右进去，然后栈里从右到左出来，这样中右左进行遍历，最后逆序一下结果，就可以变成左右中了，哈哈哈。

代码

""" # Definition for a Node. class Node: def __init__(self, val=None, children=None): self.val = val self.children = children """ class Solution: def solve(self,root,res): if(root is None): return for child in root.children: self.solve(child,res) res.append(root.val) def postorder(self, root: 'Node') -> List[int]: res=[] self.solve(root,res) return res

参考文献

[LeetCode] Beats 98.7%, 8 line pythonic iterative solution