Given the root of a binary tree, return the level order traversal of its nodes' values (left to right, level by level). Use a queue (BFS): process all nodes at the current level before moving to the next. Track level size to group nodes correctly.

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Pythontime O(n) · space O(n)

from collections import deque

def levelOrder(root) -> list[list[int]]:
    if not root:
        return []
    result, queue = [], deque([root])
    while queue:
        level = []
        for _ in range(len(queue)):
            node = queue.popleft()
            level.append(node.val)
            if node.left:  queue.append(node.left)
            if node.right: queue.append(node.right)
        result.append(level)
    return result

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