[leetcode] 589. N-ary Tree Preorder Traversal

Description

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:

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

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: [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 1000The total number of nodes is between [0, 10^4]

分析

题目的意思是：求一个n叉树的先序遍历，用递归的方法，对于当前节点的孩子节点当然是从左到右遍历了，用一个res数组在递归的过程中记录遍历的值就行了。题目问能不能用迭代的方法来实现，就需要用到队列了，从队列开头取出节点来拓展子节点放在队列末尾，然后把当前节点的值加入结果结婚就行了，直到队列的值为空为止。

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 res.append(root.val) for child in root.children: self.solve(child,res) def preorder(self, root: 'Node') -> List[int]: if(root is None): return [] res=[] self.solve(root,res) return res

参考文献

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