leetcode 116- Populating Next Right Pointers in Each Node

题目：

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

题目给出的二叉树是完美二叉树，要求是将每层的结点依次连接起来，并且每层的最后一个结点的next赋为null。

思路：
首先将每一层的最后一个结点的next赋值为null。
然后依次操作每一层，首先获取每一层的第一个结点。(其实二叉树和链表有点像，完美二叉树的最外两侧就可以看做两张链表，left和right指针类似链表的next指针)
引入两个指针，分别是cur和pre。 cur指向当前操作层，pre指向上一层。
1.cur是pre的左孩子，直接将cur的next指向pre的right结点，cur变成cur->next;
2.cur是pre的右孩子，并且pre的next不为null，cur->next指向pre->next->left;
3.cur是pre的右孩子，并且pre的next为null，说明cur已经指向了该层的最后一个结点，直接将cur->next赋为null即可。
  执行到此处，要更新pre和cur，让它们分别指向自己下一层的第一个结点。(引入计数变量count，每次执行到此处count加1，根据count更新pre和cur)
循环结束条件：cur为null。


代码：

    public void connect(TreeLinkNode root){
        if(root == null){
            return;
        }
        
        TreeLinkNode cur = root, pre = root;
        //把每层最右侧的结点的next赋为null
        while(cur != null){
            cur.next = null;
            cur = cur.right;
        }
        int count = 0;
        cur = root.left;
        while(true){
            if(cur == pre.left){
                cur.next = pre.right;
                cur = cur.next;
            }else{
                if(cur == pre.right && pre.next != null){
                    cur.next = pre.next.left;
                    cur = cur.next;
                    pre = pre.next;
                }else{
                    cur.next = null;
                    count ++;
                    pre = root;
                    for(int i = 0; i < count; i++){
                        pre = pre.left;
                    }
                    cur = pre.left;
                }
            }
        }
        
    }