Populating_Next_Right_Pointers_In_Each_Node_II.py

Given a binary tree

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.

Follow up:
You may only use constant extra space.
Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.
 
Example 1:
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.
 

Constraints:
The number of nodes in the given tree is less than 6000.
-100 <= node.val <= 100


"""
# 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
"""


# My Solution
class Solution:
    def connect(self, root: 'Node') -> 'Node':
        level = root 
        while level:
            current = level
            level = Node()
            ptr = level
            
            while current:
                if current.left:
                    level.next = current.left
                    level = level.next
                if current.right:
                    level.next = current.right
                    level = level.next
                current = current.next
                
            level = ptr.next
            
        return rootR