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描述
Given an n-ary tree, return the level order 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).
Example 1:
Input: root = [1,null,3,2,4,null,5,6]
Output: [[1],[3,2,4],[5,6]]
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],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]
Note:
The height of the n-ary tree is less than or equal to 1000
The total number of nodes is between [0, 10^4]
解析
根据题意，给定一个 n 叉树，按层级遍历顺序返回其节点值。题目中还给出了层级遍历的概念，以及输入的 N 叉树形式。
这道题其实就是考察使用层级遍历方式完成 N 叉树的层级遍历，大体思路和二叉树的层级遍历一样，我们使用一个队列 deque 来保存节点，然后使用循环进行遍历 deque 的操作，每次循环我们将当前层的节点从左到右依次弹出并放入 tmp 中进行记录，并将每个弹出节点的下一层的节点挨个依次追加到 deque 的末尾，当前层遍历结束之后将 tmp 追加到 result 中，如此循环结束之后，最后返回 result 就是按照层级遍历的结果。
时间复杂度为 O(N) ，空间复杂度为 O(N) 。
解答
class Solution(object):
def levelOrder(self, root):
"""
:type root: Node
:rtype: List[List[int]]
"""
if not root:
return []
deque = collections.deque([root])
result = []
while deque:
tmp = []
for _ in range(len(deque)):
x = deque.popleft()
tmp.append(x.val)
for node in x.children:
deque.append(node)
result.append(tmp)
return result
运行结果
Runtime: 60 ms, faster than 52.05% of Python online submissions for N-ary Tree Level Order Traversal.
Memory Usage: 16.6 MB, less than 19.56% of Python online submissions for N-ary Tree Level Order Traversal.
原题链接
https://leetcode.com/problems/n-ary-tree-level-order-traversal/
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