题目：
Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node's descendants. The tree scould also be considered as a subtree of itself.
Example 1: Given tree s:
3 / \ 4 5 / \ 1 2 Given tree t: 4 / \ 1 2 Return true , because t has the same structure and node values with a subtree of s.
Example 2: Given tree s:
3 / \ 4 5 / \ 1 2 / 0 Given tree t: 4 / \ 1 2 Return false .题目链接
题意：
给两个非空的二叉树s和t，编写函数实现判断t是否为s的子树。其中，s也可以是它自己的子树。
先序遍历二叉树s，当判断s当前结点的val和t的val相同，则进入判断子树的函数，同步遍历s和t，看是否完全相同，假如不相同，则进行下一层遍历。
代码如下：
class Solution { public: bool dfsSubtree(TreeNode* s, TreeNode * t) { // 判断两个子树是否完全相同 if ((!s && t) || (s && !t)) return false; else if (!s && !t) return true; return s->val == t->val && dfsSubtree(s->left, t->left) && dfsSubtree(s->right, t->right); } bool isSubtree(TreeNode* s, TreeNode* t) { if ((!s && t) || (s && !t)) return false; return ((s->val == t->val) && dfsSubtree(s, t) || isSubtree(s->left, t) || isSubtree(s->right, t)); // 先序遍历同时判断s->val和t->val } };