update 117 and add 117 to readme

README.md

@@ -95,6 +95,7 @@ LeetCode solutions with Chinese explanation. LeetCode中文题解。
 | 113 |[Path Sum II](https://leetcode.com/problems/path-sum-ii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/113.%20Path%20Sum%20II.md)|Medium|  |
 | 114 |[Flatten Binary Tree to Linked List](https://leetcode.com/problems/flatten-binary-tree-to-linked-list/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/114.%20Flatten%20Binary%20Tree%20to%20Linked%20List.md)|Medium|  |
 | 116 |[Populating Next Right Pointers in Each Node](https://leetcode.com/problems/populating-next-right-pointers-in-each-node/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/116.%20Populating%20Next%20Right%20Pointers%20in%20Each%20Node.md)|Medium|  |
+| 117 |[Populating Next Right Pointers in Each Node II](https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/117.%20Populating%20Next%20Right%20Pointers%20in%20Each%20Node%20II.md)|Medium|  |
 | 118 |[Pascal's Triangle](https://leetcode.com/problems/pascals-triangle)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/118.%20Pascal's%20Triangle.md)|Easy|  |
 | 119 |[Pascal's Triangle II](https://leetcode.com/problems/pascals-triangle-ii)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/119.%20Pascal's%20Triangle%20II.md)|Easy|  |
 | 121 |[Best Time to Buy and Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.md)|Easy|  |

solutions/117. Populating Next Right Pointers in Each Node II.md

@@ -4,9 +4,11 @@
 题目要求将二叉树同一层的节点用指针串起来，要求空间复杂度O(1)。
 这题是[116](https://leetcode.com/problems/populating-next-right-pointers-in-each-node/)的升级版，116中的二叉树是满二叉树，比较简单，但思路其实也是类似的，可先参考[116题解](https://github.com/ShusenTang/LeetCode/blob/master/solutions/116.%20Populating%20Next%20Right%20Pointers%20in%20Each%20Node.md)。
 ## 思路一
-最好想的还是递归: 如果root的左右子树都已经串好了的话，那么我们只需要将左子树每一层的最右边的节点的next指向右子树对应层最左边节点即可。
+最好想的还是递归: 如果root的左右子树都已经串好了的话，那么我们只需要将左子树每一层的最右边的节点(代码中用last_left表示)的next指向右子树对应层最左边节点(代码中用start_right表示)即可。          
+我们分别用start_left和start_right代表左右子树某层的最开始节点。然后我们可以从start_left一路向右(即一路next)找到左子树此层最右边那个节点last_left, 然后我们将其next指向start_right，然后更新指针start_left和start_right进入下一层。
 
 ## 思路二
+也可以用非递归的方式。我们先new一个节点head作为每一层的头结点，然后从开始从左往右从上往下遍历，然后用cur记录当前节点（初始化为head），遇到下一个节点就将cur的next指向cur，然后更新cur。
 
 # C++
 ## 思路一
@@ -14,25 +16,27 @@
 class Solution {
 public:
     Node* connect(Node* root) {
-        if(root == NULL) return NULL;
+        if(root == NULL) return NULL; // 递归出口
         Node *start_left = connect(root -> left);
         Node *start_right = connect(root -> right);
+        
         while(start_left && start_right){
             Node *last_left = start_left;
             while(last_left -> next) last_left = last_left -> next;
             last_left -> next = start_right;
             
-            while(!start_left->left && 
-                  !start_left->right && 
-                  start_left -> next != start_right) 
+            // 更新start_left
+            while(!start_left->left && // 无左子树
+                  !start_left->right && // 无右子树
+                  start_left -> next != start_right) // 不是此层最后一个节点
                 start_left = start_left -> next;
             if(start_left -> left) start_left = start_left -> left;
             else start_left = start_left -> right;
 
-            
-            while(!start_right->left && 
-                  !start_right->right && 
-                  start_right -> next) 
+            // 更新start_right
+            while(!start_right->left && // 无左子树
+                  !start_right->right && // 无右子树
+                  start_right -> next) // 不是此层最后一个节点
                 start_right = start_right -> next;
             if(start_right -> left) start_right = start_right -> left;
             else start_right = start_right -> right;
@@ -44,3 +48,30 @@ public:
 ```
 
 ## 思路二
+``` C++
+class Solution {
+public:
+    Node* connect(Node* root) {
+        Node *head = new Node(0, NULL, NULL, NULL);
+        Node *p = root, *cur = head;
+        while(p){
+            if(p -> left){
+                cur -> next = p -> left;
+                cur = cur -> next;
+            }
+            if(p -> right){
+                cur -> next = p -> right;
+                cur = cur -> next;
+            }
+            
+            p = p -> next;
+            if(!p){ // 到达一层的结尾
+                p = head -> next;
+                head -> next = NULL; // 清空头结点
+                cur = head;
+            }      
+        }
+        return root;
+    }
+};
+```