[LeetCode] 102. Binary Tree Level Order Traversal_Medium tag: BFS

Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   /   9  20
    /     15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

这个题目非常明显的适用BFS, 从root, 依次将每一层append进入ans里, 只是要注意的就是如果判断哪个是在哪一层呢, 这里用到的方法是利用size of queue, 每次得到的size就是该层的数目, 每层用一个temp list去存每一层的元素, 结束之后append进入ans中. 

1. Constraints  1) tree 可以为empty, 所以edge case 为root == None

2. Ideas

BFS   T: O(n)     S: O(n)    n is number of nodes in the tree

3. code

 1 class Solution:
 2     def LevelOrderTraversal(self, root):
 3         ans, queue = [], collections.deque([root])
 4         while queue:
 5             size, temp = len(queue) , []  # get length of the level
 6             for _ in range(size):
 7                 node = queue.popleft()
 8                 if node:
 9                     temp.append(node.val)
10                     if node.left:
11                         queue.append(node.left)
12                     if node. right:
13                         queue.append(node.right)
14             if temp:   # for edge case when root is None
15                 ans.append(temp)
16         return ans

4. Test cases

1) None   =>   []

2) [1]    =>  [[1]]

3)

Given binary tree [3,9,20,null,null,15,7],

    3
   /   9  20
    /     15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

原文地址：https://www.cnblogs.com/Johnsonxiong/p/9261520.html