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

 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) return;
12         LinkedList<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
13         queue.add(root);
14         int ParentNumInQ = 1;
15         int ChildNumInQ = 0;
16         TreeLinkNode pre = null;
17         while (!queue.isEmpty()) {
18             TreeLinkNode cur = queue.poll();
19             ParentNumInQ--;
20             if (pre == null) {
21                 pre = cur;
22             }
23             else {
24                 pre.next = cur;
25                 pre = pre.next;
26             }
27             if (cur.left != null) {
28                 queue.add(cur.left);
29                 ChildNumInQ++;
30             }
31             if (cur.right != null) {
32                 queue.add(cur.right);
33                 ChildNumInQ++;
34             }
35             if (ParentNumInQ == 0) {
36                 ParentNumInQ = ChildNumInQ;
37                 ChildNumInQ = 0;
38                 pre.next = null;
39                 pre = null;
40             }
41         }
42     }
43 }

 1 public class Solution {
 2
 3     //based on level order traversal
 4     public void connect(TreeLinkNode root) {
 5
 6         TreeLinkNode head = null; //head of the next level
 7         TreeLinkNode prev = null; //the leading node on the next level
 8         TreeLinkNode cur = root;  //current node of current level
 9
10         while (cur != null) {
11
12             while (cur != null) { //iterate on the current level
13                 //left child
14                 if (cur.left != null) {
15                     if (prev != null) {
16                         prev.next = cur.left;
17                     } else {
18                         head = cur.left;
19                     }
20                     prev = cur.left;
21                 }
22                 //right child
23                 if (cur.right != null) {
24                     if (prev != null) {
25                         prev.next = cur.right;
26                     } else {
27                         head = cur.right;
28                     }
29                     prev = cur.right;
30                 }
31                 //move to next node
32                 cur = cur.next;
33             }
34
35             //move to next level
36             cur = head;
37             head = null;
38             prev = null;
39         }
40
41     }
42 }