Peel an N-ary tree one leaf layer per day

Every day, all leaves of an N-ary tree fall off at the same time. A leaf is a node with no children left. When a node loses its last child, it becomes a leaf and falls on the next day, not the same day. This repeats until the tree is gone. The root always falls last.

Implement:

class Node:
    val: int
    children: list["Node"]

def leaves_by_day(root: Node | None) -> list[list[int]]: ...

Return days, where days[i] lists the values of the nodes removed on day i + 1. Values inside one day may be in any order. Return [] for an empty tree.

Constraints

• 0 <= n <= 10^5 nodes. The tree can be very deep (a single chain).
• Values are integers and may repeat. Treat every node as distinct.
• You do not need to change the tree. Just report the removal schedule.

Example 1

        10
      /  |  \
    20  30   40
   /  \      |
  50  60     70

Output: [[30, 50, 60, 70], [20, 40], [10]]. On day 1, 30, 50, 60 and 70 are leaves. On day 2, 20 and 40 have no children left. On day 3, only 10 is left.

Example 2

A chain 1 -> 2 -> 3 (each node has one child). Output: [[3], [2], [1]].

Reconstructed from the reported question; details may differ from what interviewers ask.