[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)