Problem URL: Populating next right pointers in each node II

We will traverse the tree with BFS and in each level, we will append the nodes to a list. After each level traversal, we take these nodes, and assing the next node to it's right node. We will repeat the process in each level and finally return our root.

"""
# 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
"""

class Solution:
    def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
        if not root:
            return None

        q = collections.deque([root])
        while q:
            level = []
            qLen = len(q)
            for _ in range(qLen):
                node = q.pop()
                level.append(node)
                if node.left: q.appendleft(node.left)
                if node.right: q.appendleft(node.right)

            for i in range(qLen-1):
                level[i].next = level[i+1]

        return root

Time Complexity: O(n)
Space Complexity: O(n)