Leetcode (Python): Populating Next Right Pointers in Each Node II

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

Solution:

# 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:
    # @param root, a tree node
    # @return nothing
    def connect(self, root):
        firstNodeLevel = root
        while firstNodeLevel != None:
            firstNodeNextLevel = None
            tempNextLevel = None
            pointer = firstNodeLevel
            while pointer != None:
                if pointer.left != None:
                    if firstNodeNextLevel == None:
                        firstNodeNextLevel = pointer.left
                    else:
                        tempNextLevel.next = pointer.left
                    tempNextLevel = pointer.left
                if pointer.right != None:
                    if firstNodeNextLevel == None:
                        firstNodeNextLevel = pointer.right
                    else:
                        tempNextLevel.next = pointer.right
                    tempNextLevel = pointer.right
                pointer = pointer.next
            firstNodeLevel = firstNodeNextLevel