[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

import collections
class Solution:
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root: return []
        ans, queue = [], collections.deque([root])
        while queue:
            size, level = len(queue), []
            for _ in range(size):
                node = queue.popleft()
                level.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            ans.append(level)
        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]
]