Populating Next Right Pointers in Each Node

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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

给定一个完全二叉树，多了一个*next变量，next指向紧邻当前节点右边的节点。我们可以用队列完成，一层一层的处理，也可以用递归完成。代码如下：
1，迭代

/**
 * 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) {
        Queue queue = new LinkedList();
        if(root == null) return;
        queue.offer(root);
        int count = 0;
        int helper = 1;
        TreeLinkNode pre = null;
        while(!queue.isEmpty()) {
            TreeLinkNode node = queue.poll();
            helper --;
            if(pre != null){
                pre.next = node;
            }
            pre = node;
            if(node.left != null) {
                queue.offer(node.left);
                count ++;
            }
            if(node.right != null) {
                queue.offer(node.right);
                count ++;
            }
            if(helper == 0) {
                pre.next = null;
                helper = count;
                count = 0;
                pre = null;
            }
        }
    }
}

2，递归

/**
 * 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;
        if(root.left != null) {
            root.left.next = root.right;
            if(root.next != null) {
                root.right.next = root.next.left;
            }
        }
        connect(root.left);
        connect(root.right);
    }
}