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

Example 2:

Input: root = [1] Output: [[1]]

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