[leetcode]Populating Next Right Pointers in Each Node II @ Python

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)