Problem

LeetCode

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]]

Constraints:

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

問題

力扣

給定一個 N 叉樹，返回其節點值的層序遍歷。（即從左到右，逐層遍歷）。

樹的序列化輸入是用層序遍歷，每組子節點都由 null 值分隔（參見示例）。

示例 1：

輸入：root = [1,null,3,2,4,null,5,6]
輸出：[[1],[3,2,4],[5,6]]

示例 2：

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

提示：

• 樹的高度不會超過 1000
• 樹的節點總數在 [0, 10^4] 之間

思路

BFS

二叉樹層次遍歷用 BFS 的時候每次都是把左右子樹加入佇列，N叉樹每次都是把所有子樹加入佇列。

Python3 程式碼

"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""

class Solution:
    def levelOrder(self, root: 'Node') -> List[List[int]]:
        import collections
        res = []
        if not root:
            return res
        
        q = collections.deque()
        q.append(root)
        # BFS
        while q:
            tmp = []
            for _ in range(len(q)):
                node = q.popleft()
                tmp.append(node.val)
                # 使用extend在列表末尾一次追加多個值
                q.extend(node.children)
            res.append(tmp)
        return res

GitHub 連結

Python