1 template<class T>
 2 class BinaryTreeNode
 3 {
 4     friend class BinaryTree<T>;
 5 private:
 6     T element;                      //结点的数据域
 7     BinaryTreeNode<T>* LeftChild;   //结点的左孩子结点
 8     BinaryTreeNode<T>* RightChild;  //结点的右孩子结点
 9 public:
10     BinaryTreeNode();
11     BinaryTreeNode(const T& ele);
12     BinaryTreeNode(const T& ele, BinaryTreeNode<T>* l, BinaryTreeNode<T>* r);
13     bool isLeaf() const;            //判断该结点是否是叶子结点，若是，则返回true
14 };

 1 template<class T>
 2 BinaryTreeNode<T>::BinaryTreeNode()
 3 {
 4     LeftChild = RightChild = NULL;
 5 }
 6 template<class T>
 7 BinaryTreeNode<T>::BinaryTreeNode(const T& ele)
 8 {
 9     element = ele;
10     LeftChild = RightChild = NULL;
11 }
12 template<class T>
13 BinaryTreeNode<T>::BinaryTreeNode(const T& ele, BinaryTreeNode<T>* l, BinaryTreeNode<T>* r)
14 {
15     element = ele;
16     LeftChild = l;
17     RightChild = r;
18 }
19 template<class T>
20 bool BinaryTreeNode<T>::isLeaf() const
21 {
22     if (LeftChild == NULL && RightChild == NULL)
23         return true;
24     else return false;
25 }

 1 template<class T>
 2 class BinaryTree
 3 {
 4 private:
 5     BinaryTreeNode<T>* root;
 6 public:
 7     BinaryTree();
 8     ~BinaryTree() {}
 9     bool IsEmpty() const;                           //判断二叉树是否为空树
10     BinaryTreeNode<T>* getRoot() const;             //返回二叉树的根结点
11     void breadthFirstOrder(BinaryTreeNode<T>* root);//广度优先遍历以root为根结点的子树
12     void preOrder(BinaryTreeNode<T>* root);         //先序遍历以root为根结点的子树
13     void inOrder(BinaryTreeNode<T>* root);          //中序遍历以root为根结点的子树
14     void postOrder(BinaryTreeNode<T>* root);        //后序遍历以root为根结点的子树
15     void deleteBinaryTree(BinaryTreeNode<T>* root); //删除以root为根结点的子树
16     void visit(BinaryTreeNode<T>* pointer);         //访问当前结点
17     BinaryTreeNode<T>* build_from_pre_and_in(char* preorder, char* inorder, int n);
18     //根据前序和中序遍历表达式构造二叉树
19     BinaryTreeNode<T>* build_from_post_and_in(char* postorder, char* inorder, int m);
20     //根据后序和中序遍历表达式构造二叉树
21     int getRootId1(char *preorder, char *inorder, int n);   //返回根结点在中序遍历表达式中序号
22     int getRootId2(char *postorder, char *inorder, int m);  //返回根结点在中序遍历表达式中序号
23 };