二叉树的类实现，为什么输出结果显示只插进去了一个结点，求大神帮忙看看我的代码哪里有问题

Question

#include
using namespace std;
//二叉树结点的ADT
template
class BinNode{
public:
E e;
BinNode* left;
BinNode* right;
BinNode(){
left=right=NULL;

}
BinNode(E ee){
e=ee;
left=right=NULL;
}
BinNode* Left(){
return left;
}
BinNode* Right(){
return right;
}
void setLeft(BinNode* l){
left=l;
}
void setRight(BinNode* r){
right=r;
}
};
//用二叉链表来实现二叉树ADT
template
class BSTNode:public BinNode{
BinNode* root;
public:
BSTNode(){
root=NULL;
}
~BSTNode(){}
BinNode* ROOT() const{ return root;}
//二叉树的构建
void create(E ee){
if(root==NULL)
root=new BinNode(ee);
else{
BinNode *temp;
BinNode *curr;
while(curr!=NULL){
temp=curr;
if(curr->e>ee)
curr=curr->left;
else
curr=curr->right;
}
if(temp->e>ee)
{temp->setLeft(new BinNode(ee));

                } 
             else
                temp->setRight(new BinNode<E>(ee));
         }
    }
    //计算二叉树结点的个数
    int NodeCount(BinNode<E> *root){
        if(root==NULL)
         return 0;
        else
         return 1+NodeCount(root->Left())+NodeCount(root->Right());
    }
    //打印结点的值
    void visit(BinNode<E> *p){
        cout<<p->e<<" ";
    } 
    //先序遍历二叉树
    void Preorder(BinNode<E> *root){
        if(root==NULL) return;
        visit(root);
        Preorder(root->left);
        Preorder(root->right);
    }
    //中序遍历二叉树
    void inorder(BinNode<E> *root){
        if(root==NULL) return;
        inorder(root->left);
        visit(root);
        inorder(root->right);
    }
    //后序遍历二叉树
    void Postorder(BinNode<E> *root){
        if(root==NULL) return;
        Postorder(root->left);
        Postorder(root->right);
        visit(root);
    }

};
int main(){
BSTNode tree1;
BSTNode tree2;
int a[9]={37,24,42,7,2,40,42,32,120};
int b[9]={120,42,42,7,2,32,37,24,40};
int i;
for(i=0;i<9;i++){
tree1.create(a[i]);
}
for(i=0;i<9;i++){
tree2.create(b[i]);
}
cout<<"第一棵二叉树的结点个数为："<<tree1.NodeCount(tree1.ROOT())<<endl;
cout<<"第一棵二叉树的先序遍历序列为：";
tree1.Preorder(tree1.ROOT());
cout<<endl;
cout<<"第一棵二叉树的中序遍历序列为：";
tree1.inorder(tree1.ROOT());
cout<<endl;
cout<<"第一棵二叉树的后序遍历序列为：";
tree1.Postorder(tree1.ROOT());
cout<<endl;
cout<<"第二棵二叉树的结点个数为："<<tree2.NodeCount(tree2.ROOT())<<endl;
cout<<"第二棵二叉树的先序遍历序列为：";
tree2.Preorder(tree2.ROOT());
cout<<endl;
cout<<"第二棵二叉树的中序遍历序列为：";
tree2.inorder(tree2.ROOT());
cout<<endl;
cout<<"第二棵二叉树的后序遍历序列为：";
tree2.Postorder(tree2.ROOT());
return 0;
}

Answer 1

temp->setLeft(new BinNode(ee));
->
curr->setLeft(new BinNode(ee));

right也一样修改