首先是自己写的迭代，用的是队列的数据结构，贴代码

 1 /*
 2 // Definition for a Node.
 3 class Node {
 4 public:
 5     int val;
 6     Node* left;
 7     Node* right;
 8     Node* next;
 9 
10     Node() : val(0), left(NULL), right(NULL), next(NULL) {}
11 
12     Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
13 
14     Node(int _val, Node* _left, Node* _right, Node* _next)
15         : val(_val), left(_left), right(_right), next(_next) {}
16 };
17 */
18 
19 class Solution {
20 public:
21     Node* connect(Node* root) 
22     {
23         if(!root)
24         return NULL;
25         int i = 1;
26         queue<Node*> Node_que;
27         Node_que.push(root);
28         while(!Node_que.empty())
29         {
30             for(int j = 0 ; j < i ; j++)
31             {
32                 Node* temp = Node_que.front();
33                 if(temp->left)
34                 {
35                     Node_que.push(temp->left);
36                     Node_que.push(temp->right);
37                 }
38                 Node_que.pop();
39                 if(j != i-1)
40                 {
41                     temp->next = Node_que.front();
42                 }
43             }
44             i = i*2;
45         }
46         return root;                
47     }
48 };

递归写不出来，还有一种迭代的思路，通过上一层实现的next指针，来实现下一层的相连，相同父节点之间的链接好办，主要是不同父节点之间的连接较难处理，在上一层有next指针的情况下，就能够有效的完成

 1 /*
 2 // Definition for a Node.
 3 class Node {
 4 public:
 5     int val;
 6     Node* left;
 7     Node* right;
 8     Node* next;
 9 
10     Node() : val(0), left(NULL), right(NULL), next(NULL) {}
11 
12     Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
13 
14     Node(int _val, Node* _left, Node* _right, Node* _next)
15         : val(_val), left(_left), right(_right), next(_next) {}
16 };
17 */
18 
19 class Solution {
20 public:
21     Node* connect(Node* root) 
22     {
23         if(!root)
24         return root;
25         Node* Left_most = root;
26         while(Left_most->left)  
27         {
28             Node* node = Left_most;
29             while(node)
30             {
31                 node->left->next = node->right;
32                 if(node->next)
33                 {
34                     node->right->next = node->next->left;
35                 }
36                 node = node->next;
37             }
38             Left_most = Left_most->left;
39         }  
40         return root;                 
41     }
42 };