Populating Next Right Pointers in Each Node II

lcpy gen -n 117   # by problem number
lcpy gen -s populating_next_right_pointers_in_each_node_ii   # by problem name

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

Input: root = [1,2,3,4,5,null,7]
Output: [1,#,2,3,#,4,5,7,#]
Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.

Input: root = []
Output: []

from __future__ import annotations


class Node:
    def __init__(
        self,
        val: int = 0,
        left: Node | None = None,
        right: Node | None = None,
        next: Node | None = None,
    ):
        self.val = val
        self.left = left
        self.right = right
        self.next = next


class Solution:
    # Time: O(n)
    # Space: O(1)
    def connect(self, root: Node | None) -> Node | None:
        current = root
        while current is not None:
            # Build the next level using the already-linked current level.
            level_head: Node | None = None
            level_tail: Node | None = None
            while current is not None:
                for child in (current.left, current.right):
                    if child is None:
                        continue
                    if level_tail is None:
                        level_head = child
                    else:
                        level_tail.next = child
                    level_tail = child
                current = current.next
            current = level_head
        return root