Populating Next Right Pointers In Each Node

lcpy gen -n 116   # by problem number
lcpy gen -s populating_next_right_pointers_in_each_node   # by problem name

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

Input: root = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
Explanation: Given the above perfect 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:
        leftmost = root
        while leftmost is not None and leftmost.left is not None:
            head = leftmost
            while head is not None:
                left = head.left
                right = head.right
                assert left is not None and right is not None
                left.next = right
                if head.next is not None:
                    right.next = head.next.left
                head = head.next
            leftmost = leftmost.left
        return root