Leetcode_num10_Populating Next Right Pointers in Each Node

题目：

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.

• 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).

先横向遍历该层的节点，若为左子树节点则指向右子树节点，右节点需在上层节点next指向不为空的情况指向下一个节点的左子树节点（内循环）

再以下一层第一个左节点为起点遍历（外循环）

</pre><pre name="code" class="python">class Solution:
    # @param root, a tree node
    # @return nothing
    def connect(self, root):
        if(root):
            pre=root
            cur=TreeNode(0)
            while(pre.left):
                cur=pre
                while(cur):
                    cur.left.next=cur.right
                    if(cur.next):
                        cur.right.next=cur.next.left
                    cur=cur.next
                pre=pre.left

Leetcode_num10_Populating Next Right Pointers in Each Node

原文：http://blog.csdn.net/eliza1130/article/details/39526945