Populating Next Right Pointers in Each Node I && II

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思路： 需要一个node记录每一层最左侧的起始点，还需要一个cur指针在每一层从左向右移动串联起next指针

 1 public void connect(TreeLinkNode root) {
 2         if (root == null) {
 3             return;
 4         }
 5         TreeLinkNode leftHead = root;
 6         while (leftHead != null && leftHead.left != null) {
 7             TreeLinkNode cur = leftHead;
 8             while (cur != null) {
 9                 cur.left.next = cur.right;
10                 cur.right.next = cur.next == null ? null : cur.next.left;
11                 cur = cur.next;
12             }
13             leftHead = leftHead.left;
14         }
15     }

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

思路：基本的思路都是一样的。但是因为这个不是个perfect binary tree. 所以在串联next指针时，就得想办法记录当前点得前一个点，我们引入pre指针。并且，每一层的起始点也不一定是该层最左侧的位置。所以用dummy node记录起点。

 1     public void connect(TreeLinkNode root) {
 2         if (root == null) {
 3             return;
 4         }
 5         TreeLinkNode leftNode = root;
 6         while (leftNode != null) {
 7             TreeLinkNode dummy = new TreeLinkNode(0);
 8             TreeLinkNode pre = dummy;
 9             TreeLinkNode cur = leftNode;
10             while (cur != null) {
11                 if (cur.left != null) {
12                     pre.next = cur.left;
13                     pre = pre.next;
14                 }
15                 if (cur.right != null) {
16                     pre.next = cur.right;
17                     pre = pre.next;
18                 }
19                 cur = cur.next;
20             }
21             leftNode = dummy.next;
22         }
23     }