原题地址：https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/
题意：
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
解题思路：和"Populating Next Right Pointers in Each Node"这道题不同的一点是，这道题的二叉树不是满的二叉树，有些节点是没有的。但是也可以按照递归的思路来完成。在编写递归的基准情况时需要将细节都考虑清楚：
代码一：
# 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):
if root:
if root.left and root.right:
root.left.next = root.right
tmp = root.next
while tmp:
if tmp.left: root.right.next = tmp.left; break
if tmp.right: root.right.next = tmp.right; break
tmp = tmp.next
elif root.left:
tmp = root.next
while tmp:
if tmp.left: root.left.next = tmp.left; break
if tmp.right: root.left.next = tmp.right; break
tmp = tmp.next
elif root.right:
tmp = root.next
while tmp:
if tmp.left: root.right.next = tmp.left; break
if tmp.right: root.right.next = tmp.right; break
tmp = tmp.next
self.connect(root.right)
self.connect(root.left)
# @connect(root.right)should be the first!!!
代码二：
思路更加精巧，代码更加简洁。
# 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):
if root:
p = root; q = None; nextNode = None
while p:
if p.left:
if q: q.next = p.left
q = p.left
if nextNode == None: nextNode = q
if p.right:
if q: q.next = p.right
q = p.right
if nextNode == None: nextNode = q
p = p.next
self.connect(nextNode)