題目：

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

　　題目的大致意思是：使用O(1)的空間對樹進行遍歷，並且設置next指針。因此不能使用遞歸和隊列進行廣度優先搜索。
　　解題思路: 采用指針，進行二叉樹的廣度優先搜索。總體是設置3個指針，分別為parent, pre, next。 
其中parent執行當前要設置next指針的上層節點。pre指向當前要設置next指針的節點， next指向下層的開始節點(因為不是完全二叉樹)
 

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * public class TreeLinkNode {
 4  *     int val;
 5  *     TreeLinkNode left, right, next;
 6  *     TreeLinkNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public void connect(TreeLinkNode root) {
11          if
 
 (root == null) {
12             return;
13         }
14         TreeLinkNode pre ;//用於設置next節點
15         TreeLinkNode next ; //標記下一行的開始節點
16         TreeLinkNode parent = root;//
17         
18          while(parent != null) {
19             pre = null;
20             next = null;
21             while(parent != null) {//對一行設置next值
22                 if (next == null) { //找到parent的下一行第一個節點
23                     next = (parent.left != null) ? parent.left : parent.right;
24                 }
25                 
26                 if (parent.left != null) { //parent存在子節點
27                     if (pre == null) {　　//如果pre為空，說明parent.left為下一行的第一個節點，因此讓pre指向它
28                         pre = parent.left;
29                     }else{　　//pre不為空，設置pre的next指針
30                         pre.next = parent.left; 
31                         pre = pre.next;
32                     }
33                 }
34                 if (parent.right != null) {
35                     if (pre == null) {
36                         pre = parent.right;
37                     }else{
38                         pre.next = parent.right;
39                         pre = pre.next;
40                     }
41                 }
42                 
43                 parent = parent.next;//遍歷parent的下一個節點
44             }
45             parent = next;//parent指向下一行的開始節點
46         }
47     }
48 }

Populating Next Right Pointers in Each Node II