116. 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

题意比较难理解。现在定义的树链表节点，多了一个next节点定义 next是值本节点同层的相邻节点

算法实现之后变成 每一层的节点用一个链表串起来，形成一个单链表

public class Solution {    public void connect(TreeLinkNode root) {        if(root==null) return;        TreeLinkNode parent=root;        TreeLinkNode next=parent.left;        while(parent!=null&&next!=null){            TreeLinkNode prev=null;            while(parent!=null){                if(prev==null){                    prev=parent.left;                }                else {                    prev.next=parent.left;                    prev=prev.next;                }                prev.next=parent.right;                prev=prev.next;                parent=parent.next;            }            parent=next;            next=parent.left;        }    }}