# [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); } }; ```