26. 树的子结构

class Solution {
public:
    bool check(TreeNode* pRoot1, TreeNode* pRoot2){
        if(pRoot2 == nullptr){
            return true;
        }
        if(pRoot1 == nullptr){
            return false;
        }
        if(pRoot1->val != pRoot2->val){
            return false;
        }else {
            return check(pRoot1->left, pRoot2->left) && check(pRoot1->right, pRoot2->right);
        }
    }
    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2) {
        //因为接下来要访问pRoot1->val和pRoot2->val, 所以提前处理nullptr的情况
        if(pRoot1 == nullptr || pRoot2 == nullptr) return false;
        if(pRoot1->val == pRoot2->val && check(pRoot1, pRoot2)){
            return true;
        }
        return HasSubtree(pRoot1->left, pRoot2) || HasSubtree(pRoot1->right, pRoot2);
    }
};