Populating Next Right Pointers in Each Node II

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

我参考的：http://zjalgorithm.blogspot.com/2014/11/leetcode-java-populating-next-right.html

思路：constant space，还是跟上一题一样，从上到下，每层扫描并连接。

这题并不是什么先处理右边再处理左边，就是从上倒下，从左到右处理，只是用pre来记录要连接的前面一个点，如何赋值pre就要分pre == null和pre ！= null两种情况讨论。

这题跟上面的I很类似，也是每层访问，从左到右进行连接，访问下一层的时候，需要注意，因为这个时候，并不是满树，而是分左右节点为空，所以下一层需要额外讨论一下。这题需要注意的是，找下一层的开始点的时候，需要在内层循环里面找，因为可能上一层最左边的点的左右子树都是空的，但是后面的node有左右子树，所以需要在内部循环里面找下一层的起点。

/**  * 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) {         if(root == null) return;         TreeLinkNode cur = root;         while(cur!=null){             TreeLinkNode nextstart = null;             TreeLinkNode pre = null;             TreeLinkNode curlevel = cur;             while(curlevel!=null){                 if(nextstart == null){                     nextstart = curlevel.left == null ? curlevel.right : curlevel.left;                 }                  if(curlevel.left!=null){                     if(pre == null){                         pre = curlevel.left;                     } else {                         pre.next = curlevel.left;                         pre = curlevel.left;                     }                 }                 if(curlevel.right!=null){                     if(pre == null){                         pre = curlevel.right;                     } else {                         pre.next = curlevel.right;                         pre = curlevel.right;                     }                 }                 curlevel = curlevel.next;             }             cur = nextstart;         }     } }