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

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题目难度为中等。

题目要求填充每一个node的next值，指向和它一层的下一个节点，并且指明了这是一个完美的二叉树，每个父节点都有两个孩子，所有叶结点都在同一层，这就大大简化了解题过程。

解题思路：运用循环来解题，由于next的存在，做起来还是很简单的，假设curNode在第n层，perNode就在n-1层，它们都是最左边的node，这个时候只需要把curNode一步一步的往右走，在走的同时填充next属性即可，LR变量用来表示该curNode是左孩子还是右孩子，当它是左孩子时，preNode也要往右走

/**
 * 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;
        }
        root.next = null;
        TreeLinkNode preNode = root;
        TreeLinkNode curNode = root.left;
        while (curNode != null) {
            TreeLinkNode temp = curNode; //备份
            int LR = 0; //0:L, 1:R
            while(true){
                if (LR == 0) { //当前节点是左孩子
                    curNode.next = preNode.right;
                    curNode = preNode.right;        
                    preNode = preNode.next;
                    LR = 1;
                } else {  //当前节点是右孩子
                    if (preNode == null) {
                        curNode.next = null;
                        break;
                    }
                    curNode.next = preNode.left;
                    curNode = preNode.left;
                    LR = 0;
                }
            }
            //进入下一层循环
            preNode = temp;
            curNode = temp.left;
        }
    }
}