add 572

Author: ShusenTang

README.md

@@ -284,6 +284,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 532 |[K-diff Pairs in an Array](https://leetcode.com/problems/k-diff-pairs-in-an-array)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/532.%20K-diff%20Pairs%20in%20an%20Array.md)|Easy|  |
 | 561 |[Array Partition I](https://leetcode.com/problems/array-partition-i)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/561.%20Array%20Partition%20I.md)|Easy|  |
 | 566 |[Reshape the Matrix](https://leetcode.com/problems/reshape-the-matrix)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/566.%20Reshape%20the%20Matrix.md)|Easy|  |
+| 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|  |
 | 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|  |

solutions/572. Subtree of Another Tree.md

@@ -0,0 +1,26 @@
+# [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)
+
+# 思路
+让我们求一个数是否是另一个树的子树，从题目中的第二个例子中可以看出，中间某个部分的不能算是子树。如果从s的某个结点开始，跟t的所有结构都一样就可以了，所以问题转换为判断两树是否相等，即[100. Same Tree](100.%20Same%20Tree.md)。
+
+
+# C++
+
+``` C++
+class Solution {
+private:
+    bool isSameTree(TreeNode* t1, TreeNode* t2){
+        if(!t1 || !t2) return (!t1 && !t2);
+        return (t1 -> val == t2 -> val) && \
+               isSameTree(t1 -> left, t2 -> left) && \
+               isSameTree(t1 -> right, t2 -> right);
+    }
+public:
+    bool isSubtree(TreeNode* s, TreeNode* t) {
+        if(!s) return t == NULL;
+        if(!t) return true;
+        return isSameTree(s, t) || isSubtree(s -> left, t) || isSubtree(s -> right, t);
+    }
+};
+```
+