動機

複習BFS

Problem

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 Leetcode 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]]

 

Constraints:

    t
  • The height of the n-ary tree is less than or equal to 1000
    • t
  • The total number of nodes is between [0, 104]

Sol

class Solution:
    def levelOrder(self, root: 'Node') -> List[List[int]]:
        if not root:
            return []
        q = deque([[root]])
        ret = []
        while q:
            rs = q.popleft()
            ret.append([r.val for r in rs])
            rs = sum([[x for x in r.children if x] for r in rs if r],[])
            if rs:
                q.append(rs)
        return ret