 6/22/2017 - 7:28 AM

## 117. Populating Next Right Pointers in Each Node II

1. Populating Next Right Pointers in Each Node II
``````# For our current solution, it doesn't matter if the tree is perfect or not.

# Definition for binary tree with next pointer.
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#         self.next = None

class Solution:
# @param root, a tree link node
# @return nothing
def connect(self, root):
if root == None:
return
queue = [root]
while len(queue) != 0:
levelNum = len(queue)
prev = None
for i in range(levelNum):
curNode = queue
del queue
curNode.next = prev
prev = curNode
if curNode.right:
queue.append(curNode.right)
if curNode.left:
queue.append(curNode.left)``````

https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/#/description

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
``````