判断R2是否为R1的子结构，递归实现

bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
    {
    

        if(pRoot2==NULL)
            return 1;
            if(pRoot1 == NULL)
            return 0;
        
        if(pRoot1->val == pRoot2->val){
            if(HasSubtree(pRoot1->left,pRoot2->left) && HasSubtree(pRoot1->right,pRoot2->right))
                {
                    return 1;
                }
            }
 
            if(HasSubtree(pRoot1->left,pRoot2) ||
                HasSubtree(pRoot1->right,pRoot2)){
                return 1;
            }
        
        
        return 0;
        
    }