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

一刷
題解：
創建3個節點，cur(處于當前層), prev(處于操作層), head(下一層的最外節點)。即 cur-1, prev-2, head-3

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode head = null;//the head of the next level
        TreeLinkNode prev = null;// the leading node of next level
        TreeLinkNode cur = root; //current node of current level
        
        //cur-1, prev-2, head-3
        
        while(cur!=null){
            while(cur!=null){
                //current manipulate-level 2
                if(cur.left!=null){
                    if(prev == null){
                        head = cur.left;
                    }
                    else prev.next = cur.left;
                    prev = cur.left;
                }
                
                if(cur.right!=null){
                    if(prev == null) head = cur.right;
                    else prev.next = cur.right;
                    prev = cur.right;
                }
                cur = cur.next;
            }
            cur = head;
            head = null;
            prev = null;
        }
    }
}