update 114 and add 114 to readme

README.md

@@ -93,6 +93,7 @@ LeetCode solutions with Chinese explanation. LeetCode中文题解。
 | 111 |[Minimum Depth of Binary Tree](https://leetcode.com/problems/minimum-depth-of-binary-tree)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/111.%20Minimum%20Depth%20of%20Binary%20Tree.md)|Easy|  |
 | 112 |[Path Sum](https://leetcode.com/problems/path-sum)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/112.%20Path%20Sum.md)|Easy|  |
 | 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|  |
 | 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/114. Flatten Binary Tree to Linked List.md

@@ -5,7 +5,7 @@
 ## 思路一
 最容易想到的就是先序遍历一遍树。我们可以用一个全局的指针`list_last`记录当前链表的最后一个节点，每遍历到一个节点就将其接到`list_last`后面（右指针）。
 
-## 思路二
+## 思路二*
 此题也可以直接递归来做。即如果root的左右子树都已经是被展开好的，那么只需要将左子树接在root的右边，再将右子树接在原来左子树的最下边就可以了，如下例。
 ```
   1               1
@@ -21,6 +21,42 @@
                             6
 ```
 
+## 思路三*
+可以将思路二改成非递归版本。原理类似：当前工作指针p指向root，如果p有左子树，那么将其左子树接在p的右边，而原来的右子树接在左子树的最右下边节点，此为一次循环，然后工作指针p向右下移动一步，重复上述操作。下面举个例子：
+```
+原二叉树:
+    1
+   / \
+  2   5
+ / \   \
+3   4   6
+
+第一次工作指针p指向1，循环后:
+   1
+    \
+     2
+    / \
+   3   4
+        \
+         5
+          \
+           6
+
+第二次工作指针p指向2，循环后:  
+   1
+    \
+     2
+      \
+       3
+        \
+         4
+          \ 
+           5
+            \
+             6
+接下来每次循环p依次指向3、4、5、6，由于p没有左子树，所以不作任何操作。
+```
+
 # C++
 ## 思路一
 ``` C++
@@ -73,4 +109,27 @@ public:
         tmp -> right = right;
     }
 };
+```
+
+## 思路三
+``` C++
+class Solution {
+public:
+    void flatten(TreeNode* root) {
+        TreeNode *p = root;
+        while(p){
+            if(p -> left){
+                // 先备份右子树
+                TreeNode *right_bk = p -> right;
+                p -> right = p -> left;
+                p -> left = NULL;
+                TreeNode *tmp = p;
+                while(tmp -> right) tmp = tmp -> right;
+                // 此时tmp指向p原来左子树的最右下节点
+                tmp -> right = right_bk;
+            }
+            p = p -> right;
+        }
+    }
+};
 ```