[Leetcode]Populating Next Right Pointers in Each Node

Given a binary tree

struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to 

NULL

.

Initially, all next pointers are set to 

NULL

.

Note:
You may only use constant extra space.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,

Given the following perfect binary tree,

1
/  \
2    3
/ \  / \
4  5  6  7

After calling your function, the tree should look like:

1 -> NULL
/  \
2 -> 3 -> NULL
/ \  / \
4->5->6->7 -> NULL

把树的每一层都维护成一个链表~假设树是完全二叉树~题目要求只能用常数空间，所以不能用深度搜索，因为递归需要用栈；也不能用广度搜索，因为队列也需要占用空间~下面接法时间复杂度为O(n)，空间复杂度为O(1)~

class Solution:
# @param root, a tree node
# @return nothing
def connect(self, root):
if root is None: return
while root.left:
curLevel = root
while curLevel:
curLevel.left.next = curLevel.right
if curLevel.next:
curLevel.right.next = curLevel.next.left
curLevel = curLevel.next
root = root.left

再附上递归解法~代码如下~

class Solution:
# @param root, a tree node
# @return nothing
def connect(self, root):
if root is None: return
if root.left and root.right:
root.left.next = root.right
if root.right and root.next:
root.right.next= root.next.left
self.connect(root.left)
self.connect(root.right)