add 617. Merge Two Binary Trees 🍺

Author: ShusenTang

README.md

@@ -321,6 +321,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 572 |[Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)|[C++](solutions/572.%20Subtree%20of%20Another%20Tree.md)|Easy|  |
 | 581 |[Shortest Unsorted Continuous Subarray](https://leetcode.com/problems/shortest-unsorted-continuous-subarray)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/581.%20Shortest%20Unsorted%20Continuous%20Subarray.md)|Easy|  |
 | 605 |[Can Place Flowers](https://leetcode.com/problems/can-place-flowers)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/605.%20Can%20Place%20Flowers.md)|Easy|  |
+| 617 |[Merge Two Binary Trees](https://leetcode.com/problems/merge-two-binary-trees/)|[C++](solutions/617.%20Merge%20Two%20Binary%20Trees.md)|Easy|  |
 | 628 |[Maximum Product of Three Numbers](https://leetcode.com/problems/maximum-product-of-three-numbers)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/628.%20Maximum%20Product%20of%20Three%20Numbers.md)|Easy|  |
 | 643 |[Maximum Average Subarray I](https://leetcode.com/problems/maximum-average-subarray-i)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/643.%20Maximum%20Average%20Subarray%20I.md)|Easy|  |
 | 661 |[Image Smoother](https://leetcode.com/problems/image-smoother)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/661.%20Image%20Smoother.md)|Easy|  |

solutions/617. Merge Two Binary Trees.md

@@ -0,0 +1,24 @@
+# [617. Merge Two Binary Trees](https://leetcode.com/problems/merge-two-binary-trees/)
+
+# 思路
+
+合并两个二叉树。就是一个很简单的递归：
+* 若`t1`和`t2`任意一个为空，那么直接返回另一个即可（递归出口）；
+* 否则，将`t1`的值加上`t2`的值，然后递归合并`t1`和`t2`的左子树以及`t1`和`t2`的右子树。
+
+# C++
+``` C++
+class Solution {
+public:
+    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
+        if(t1 == NULL) return t2;
+        if(t2 == NULL) return t1;
+        
+        t1 -> val += t2 -> val;
+        t1 -> left = mergeTrees(t1 -> left, t2 -> left);
+        t1 -> right = mergeTrees(t1 -> right, t2 -> right);
+        delete t2;
+        return t1;
+    }
+};
+```