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

/**
 * 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;
        Queue<TreeLinkNode> queue = new LinkedList<>();
        queue.offer(root);
        TreeLinkNode pre = null;
        while (!queue.isEmpty()) {
            int size = queue.size();
           for (int i=0; i<size; i++) {
                TreeLinkNode cur = queue.poll();
                if (i == 0) pre = cur;
                else {
                    pre.next = cur;
                    pre = cur;
                }

                
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
            }
            
        }
    }
}

public class Solution {
    public void connect(TreeLinkNode root) {
        if(root==null) return;
		root.next = null;
        TreeLinkNode cur = root;
        TreeLinkNode nextLeftmost = null;

      //递进while(cur.left!=null){
        //递进nextLeftmost = cur.left; // save the start of next level, 当作下次内循环的开始点并判断是否有下次循环
       //层次  while(cur!=null){
                cur.left.next=cur.right;
                cur.right.next = cur.next==null? null : cur.next.left;
                cur=cur.next;
            }
            cur=nextLeftmost;  // point to next level 
        }
    }
}