这个二叉树的功能算很全面了~~
由给定的完全二叉树形式存储的数组(如"12345 6"),构造二叉树
提供:复制构造函数和赋值操作符重载,递归和非递归形式的中、前、后序遍历方法,求一个节点的父节点,左右兄弟结点的函数以及 求二叉树深度和结点个数的函数。
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// BTreeNode.h
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// 二叉树结点抽象类型
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#ifndef BTREENODE_H
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#define BTREENODE_H
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#include <cstdlib>
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template<class T> class BTree;
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template<class T> class BTreeNode
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{
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friend class BTree<T>;
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public:
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BTreeNode():lchild(NULL),rchild(NULL)
{ };
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BTreeNode(const T&dt, BTreeNode<T> *lch =NULL , BTreeNode<T> *rch = NULL)
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:data(dt),lchild(lch),rchild(rch)
{};
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T get_data()const
{return data; };
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BTreeNode<T>* get_lchild()const
{return lchild; };
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BTreeNode<T>* get_rchild()const
{return rchild; };
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private:
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T data;
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BTreeNode<T> *lchild, *rchild;
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};
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#endif
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/**//************************************************************************
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** BTree.h二叉树抽象类型
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** 由给定的完全二叉树形式存储的数组(如"12345 6"),构造二叉树
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** 提供:复制构造函数和赋值操作符重载
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** 递归和非递归形式的中、前、后序遍历方法
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** 求一个节点的父节点,左右兄弟结点的函数
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** 求二叉树深度和结点个数的函数
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************************************************************************/
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#ifndef BTREE_H
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#define BTREE_H
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#include "BTreeNode.h"
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#include <stack> //非递归遍历时借用栈
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#include <cstdlib> //NULL的定义在这里
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template <class T>
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class BTree
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{
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private:
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BTreeNode<T>* build_body(T*elems, int n, int i); //构造二叉树时使用
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BTreeNode<T>* root;
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public:
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//三个构造函数
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BTree(T *a,int m);
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BTree(BTreeNode<T> *p = NULL)
{ root = new BTreeNode<T>; copy_body(root, p); };
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BTree(const T& t, BTree<T>& ltree, BTree<T>& rtree)
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{
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root = new BTreeNode<T>(t, ltree.root, rtree.root);
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//原来两颗子树的根结点设置为空,避免非法访问,否则遍历时会出错,创建新树之前可以复制原来两棵树
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ltree.root = rtree.root = NULL;
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};
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//复制构造函数
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BTree(BTree& bt)
{root = new BTreeNode<T>; copy_body(root,bt.root);};
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~BTree()
{ destry(root); }
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//重载复制操作符
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BTree<T>& operator = (const BTree<T>& nbt)
{root = new BTreeNode<T>; copy_body(root,nbt.root);};
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//递归复制二叉树
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static void copy_body(BTreeNode<T>*&p, BTreeNode<T>* q);
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//递归遍历
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static void in_order(BTreeNode<T>*p, void visit(BTreeNode<T>* p));
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static void pre_order(BTreeNode<T>*p, void visit(BTreeNode<T>* p));
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static void post_order(BTreeNode<T>*p, void visit(BTreeNode<T>* p));
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virtual void pre_order(void visit(BTreeNode<T>* p))const
{pre_order(root, visit);};
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virtual void in_order(void visit(BTreeNode<T>* p))const
{in_order(root, visit); };
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virtual void post_order(void visit(BTreeNode<T>* p))const
{post_order(root, visit); };
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//非递归遍历
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virtual void in_order1(void visit(BTreeNode<T>* p))const;
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virtual void pre_order1(void visit(BTreeNode<T>* p))const;
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virtual void post_order1(void visit(BTreeNode<T>* p))const;
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BTreeNode<T>* get_root()const
{return root;}; //返回二叉树根
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BTreeNode<T>* get_parent(BTreeNode<T> *curr)const; //返回给定结点父节点
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//定义见get_parent函数,只需修改一条语句
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BTreeNode<T>* get_left_sibling(BTreeNode<T>* curr)const; //返回给定结点左兄弟结点
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//定义见get_parent函数,只需修改一条语句
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BTreeNode<T>* get_right_sibling(BTreeNode<T>* curr)const; //返回给定结点右兄弟结点
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void set_root(BTreeNode<T>* p)
{ destry(root); copy_body(root, p);};
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//释放内存资源
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static void destry(BTreeNode<T> *&p);
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//求二叉树结点个数
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static int num(BTreeNode<T>* p);
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int num()const
{ return num(root);};
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//求二叉树深度
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static int depth(BTreeNode<T>* p);
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int depth()const
{return depth(root);};
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//判断二叉树是否相等
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static bool equal(BTreeNode<T> *p, BTreeNode<T> *q);
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bool operator ==(BTree<T>&bt) const
{return equal(root, bt.root);};
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};
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//构造函数
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template<class T>
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BTree<T>::BTree(T *a,int m)
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{
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root = build_body(a, m, 1);//自作聪明,从0开始调用函数,导致l=2*i=0,没有输出结果,莫名其妙了半天
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};
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//构造子树
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template <class T>
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BTreeNode<T>* BTree<T>::build_body(T*elems, int n, int i)//suffix
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{
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BTreeNode<T> *p;
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int l,r;
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if( i <= n && elems[i-1] != ' ')
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{
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p = new BTreeNode<T>;
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p->data = elems[i-1];
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l = 2*i; //左儿子结点位置
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r = 2*i + 1; //右儿子结点位置
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p->lchild = build_body(elems,n,l);
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p->rchild = build_body(elems,n,r);
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return p;
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}
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else
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return NULL;
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}
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//复制二叉树
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template<class T>
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void BTree<T>::copy_body(BTreeNode<T>* &p, BTreeNode<T> *q)
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{
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if(q != NULL)
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{
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if(p == NULL)
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p = new BTreeNode<T>;
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p->data = q->data;
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copy_body(p->lchild,q->lchild);
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copy_body(p->rchild,q->rchild);
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}
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else p = NULL;
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}
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//递归中序遍历
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template<class T>
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void BTree<T>::in_order(BTreeNode<T>*p, void visit(BTreeNode<T>* p))
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{
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if(p != NULL)
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{
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in_order(p->lchild,visit);
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visit(p);
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in_order(p->rchild,visit);
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}
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}
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//递归前序遍历
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template<class T>
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void BTree<T>::pre_order(BTreeNode<T>*p,void visit(BTreeNode<T>*p))
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{
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if(p != NULL)
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{
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visit(p);
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pre_order(p->lchild,visit);
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pre_order(p->rchild,visit);
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}
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}
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//递归后序遍历
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template<class T>
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void BTree<T>::post_order(BTreeNode<T>*p,void visit(BTreeNode<T>*p))
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{
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if(p != NULL)
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{
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post_order(p->lchild,visit);
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post_order(p->rchild,visit);
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visit(p);
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}
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}
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//非递归中序遍历
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template<class T>
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void BTree<T>::in_order1( void visit(BTreeNode<T>* p))const
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{
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cout << "The inorder is : ";
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std::stack< BTreeNode<T>* > stk;
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BTreeNode<T> * q;
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stk.push(root); //根结点进栈
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while( !stk.empty()) //若栈非空,重复
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{
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while(stk.top() != NULL) //栈顶结点的左儿子相继进栈,直到NULL为止
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{
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q= stk.top()->lchild;
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stk.push(q);
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}
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stk.pop(); //将NULL退出栈
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if( !stk.empty()) //访问栈顶结点,并将其跳出栈
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{
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q = stk.top();
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stk.pop();
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visit(q);
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stk.push(q->rchild); //将原栈顶结点的右子树压入栈
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}
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}
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cout << endl;
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}
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//非递归前序遍历
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template<class T>
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void BTree<T>::pre_order1(void visit(BTreeNode<T>*p))const
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{
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cout << "The preorder is: " ;
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std::stack< BTreeNode<T>* > stk;
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BTreeNode<T>* q;
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visit(root);
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stk.push(root); //访问根节点,并将其压入栈
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while( !stk.empty())
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{
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while(stk.top() != NULL) //相继访问栈顶结点的左儿子,并将其压入栈,直至NULL
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{
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q = stk.top()->lchild;
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if(q != NULL) visit(q);
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stk.push(q);
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}
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stk.pop(); // 将NULL跳出栈
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if( !stk.empty())
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{
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q = stk.top()->rchild; //标记原栈顶结点右儿子结点
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stk.pop(); // 栈顶结点跳出栈
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if( q != NULL) visit(q);
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stk.push(q); //访问右儿子结点并将其压入栈
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}
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}
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cout << endl;
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}
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//非递归后序遍历
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template <class T>
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void BTree<T>::post_order1(void visit(BTreeNode<T>* p))const
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{
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cout << "The postorder is: ";
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std::stack< BTreeNode<T>* > stk;
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BTreeNode<T> *q = NULL, *pre = NULL; //pre记录前一个访问的节点
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stk.push(root);
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while( !stk.empty()) //最后是树根是如何跳出循环
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{
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while(stk.top() != NULL) //栈顶结点的左儿子结点相继进栈,直到NULL
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stk.push(stk.top()->lchild);
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stk.pop(); //NULL跳出栈
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if(!stk.empty())
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{
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q = stk.top(); //栈顶结点跳出
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stk.pop();
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// 当栈顶结点有右儿子,且没有被访问过时,将原栈顶结点重新压入栈,并将其右儿子也压入栈
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if(q->rchild != NULL && !equal(q->rchild,pre))
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{
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stk.push(q);
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stk.push(q->rchild);
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}
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// 不然,访问原栈顶结点,为防止重复遍历右儿子,将NULL压入栈
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else
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{
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visit(q);
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stk.push(NULL);
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pre = q;
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}
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}
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}
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cout << endl;
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}
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//求双亲结点
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//实质就是在中序遍历的时候顺便判断一下给定结点是不是某一节点的儿子结点
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template<class T>
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BTreeNode<T>* BTree<T>::get_parent(BTreeNode<T> *curr)const
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{
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cout << "The parent of " << curr->get_data() << " is: ";
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std::stack< BTreeNode<T>* > stk;
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BTreeNode<T> *p, *q;
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p = NULL;
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stk.push(root);
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while( !stk.empty())
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{
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while(stk.top() != NULL)
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{
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q= stk.top()->lchild;
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stk.push(q);
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}
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stk.pop();
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if( !stk.empty())
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{
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q = stk.top();
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stk.pop();
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//求双亲结点
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if(q->lchild == curr || q->rchild == curr) p = q;
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//求左兄弟结点
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//if(q->rchild == curr) p = q->lchild;
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//求右兄弟结点
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//if(q->lchild == curr) p = q->rchild;
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stk.push(q->rchild);
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}
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}
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return p;
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}
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//释放资源
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template<class T>
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void BTree<T>::destry(BTreeNode<T> *&p)
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{
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if(p != NULL)
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{
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destry(p->lchild);
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destry(p->rchild);
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delete p;
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}
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p = NULL;
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}
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//求二叉树结点个数
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template<class T>
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int BTree<T>::num(BTreeNode<T>* p)
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{
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if(p == NULL) return 0;
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else return num(p->lchild) + num(p->rchild) + 1;
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}
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//求二叉树深度
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template<class T>
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int BTree<T>::depth(BTreeNode<T>* p)
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{
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int max = 0;
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if(p == NULL) return 0;
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else
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{
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max = depth(p->lchild);
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if(depth(p->rchild) > max) max = depth(p->rchild);
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return (max + 1);
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}
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}
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//判断二叉树是否相等
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template<class T>
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bool BTree<T>::equal(BTreeNode<T> *p, BTreeNode<T> *q)
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{
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bool b = true;
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if((p == NULL) && (q == NULL)) ;
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else if((p == NULL) || (q == NULL)) b = false;
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else
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{
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b = (p->data == q->data) && (p->lchild == q->lchild) && (p->rchild == q->rchild);
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}
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return b;
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}
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#endif
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//MainFn.cpp 测试二叉树功能
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#include "BTree.h"
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#include <iostream>
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using std::endl;
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using std::cout;
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//谓词函数predicate,定义访问二叉树结点的形式
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void visit(BTreeNode<char> *p)
{ cout << p->get_data() << " ";};
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int main()
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{
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char *str = "12345 6";//字符数组的定义形式
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char *str2 = "78 9";
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BTree<char> bt(str,7);
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BTree<char> bt_copy(bt);
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BTree<char> bt_copy2(bt.get_root());
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BTree<char> bt_copy3 = bt;
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BTree<char> bt2(str2,4);
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//BTree<char> bt3('a',bt,bt2); //创建新树,注意创建完这个树以后原先的两个树bt,bt2已经无效,不能再调用
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//bt3.in_order1(visit);
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//测试构造函数
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bt.in_order1(visit);
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bt_copy.in_order1(visit);
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bt_copy2.in_order1(visit);
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bt_copy3.in_order1(visit);
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//测试遍历函数
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bt.pre_order1(visit);
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cout << "Postorder is :";
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bt.post_order(visit);
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cout <<endl;
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bt.post_order1(visit);
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//测试求二叉树结点个数和深度的函数及其他
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cout << bt.num() << "," << bt.depth() <<endl;
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cout << bt.get_parent(bt.get_root()->get_lchild())->get_data() <<endl; //求双亲结点
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return 0;
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}