BFS level-order traversal

Queue-driven level-by-level visit.

BFS trees queues

Implement:
```
class TreeNode:
    def __init__(self, val, left=None, right=None): ...

def level_order(root: TreeNode | None) -> list[list[int]]:
```
Return `[]` for an empty tree.

Constraints

0 ≤ nodes ≤ 2000

Example

root = TreeNode(3, TreeNode(9), TreeNode(20, TreeNode(15), TreeNode(7)))
level_order(root)  # [[3],[9,20],[15,7]]

28 points ~25 min

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Hints

Use a deque; enqueue children; track level boundaries.

Test Results

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