LeetCode 429 - N-ary Tree Level Order Traversal (Medium)

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, 10^4]

 方法：BFS 层遍历就可以。但是要注意当root为空时候的return，应该是[]，而不是[[]]

time  complexity: O(n) space  complexity: O(n) 

"""
# 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]]:
        if not root: return []
        
        res = []
        q = collections.deque()
        q.append(root)
        
        while q:
            tmp = []
            for _ in range(len(q)):
                node = q.pop()
                if node:
                    tmp.append(node.val)
                    for child in node.children:
                        q.appendleft(child)
            if tmp:
                res.append(tmp)
        
        return res