 # Problem

Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).

Example 1:

Input: root = [3,9,20,null,null,15,7] Output: [,[9,20],[15,7]]

Example 2:

Input: root =  Output: []

Example 3:

Input: root = [] Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -1000 <= Node.val <= 1000

# Solution

## With recursion:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
levels = []
if not root:
return levels

def helper(node, level):
# start the current level
if len(levels) == level:
levels.append([])

# append the current node value
levels[level].append(node.val)

# process child nodes for the next level
if node.left:
helper(node.left, level + 1)
if node.right:
helper(node.right, level + 1)

helper(root, 0)
return levels

This is a bit confusing and I don't like it and I didn't come up with it. I'd prefer a BFS method with a queue.

## BFS

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
res = []
# queue
q = collections.deque()
q.append(root)

while q:
# while queue is non-empt
qLen = len(q)
# loop through all values in the queue
# makes sure we iterate one level at a time
level = []
# for all nodes in that level
for i in range(qLen):
node = q.popleft()
# FIFO by popping left
if node:
level.append(node.val)
# technically these could be null, hence the if node above
q.append(node.left)
q.append(node.right)
if level:
res.append(level)
return res