# You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:
#
# struct Node {
#   int val;
#   Node *left;
#   Node *right;
#   Node *next;
# }
# Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
#
# Initially, all next pointers are set to NULL.
#
#
#
# Example 1:
#
#
# 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.
# Example 2:
#
# Input: root = []
# Output: []
#
#
# Constraints:
#
# The number of nodes in the tree is in the range [0, 212 - 1].
# -1000 <= Node.val <= 1000
#
#
# Follow-up:
#
# You may only use constant extra space.
# The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.


"""
# Definition for a Node.
class Node:
    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""

from collections import deque


class Solution:
    def connect(self, root: 'Node') -> 'Node':

        if not root:  # check if tree empty
            return root

        queue = deque([root, ])

        while queue:
            # start the level array
            n = len(queue)

            for i in range(n):
                node = queue.popleft()  # get node from start of Q

                if i + 1 != n: # SINCE BALANCED TREE 2 ELES AND THUS ONE IN QUEUE IS NEXT ELEMENT
                    node.next = queue[0]
                # NO NEED TO CHECK NEXT AS BY DEFAULT IT WILL BE NONE/NULL

                if node.left:  # left tree from pop
                    queue.append(node.left)
                if node.right:  # right tree from pop
                    queue.append(node.right)

        return root