117 Populating Next Right Pointers in Each Node II.py

"""
Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.
For example,
Given the following binary tree,
         1
       /  \
      2    3
     / \    \
    4   5    7
After calling your function, the tree should look like:
         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL
Author: Rajeev Ranjan
"""

# Definition for a  binary tree node
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
        self.next = None


class Solution:
    def connect(self, root):
        """
        bfs
        same as Populating Next Right Pointers in Each Node I
        :param root: TreeNode
        :return: nothing
        """
        if not root:
            return None

        q = [root]
        while q:
            current_level = q
            q = []
            for ind, val in enumerate(current_level):
                if val.left: q.append(val.left)
                if val.right: q.append(val.right)
                try:
                    val.next = current_level[ind+1]
                except IndexError:
                    val.next = None