第十周上机实践—项目3—利用二叉遍历思想解决问题

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/* *Copyright(c) 2015,烟台大学计算机学院 *All rights reserved. *文件名称:test.cpp *作者:林莉 *完成日期:2015年9月11日 *版本:v1.0 * *问题描述:假设二叉树采用二叉链存储结构存储，分别实现以下算法，并在程序中完成测试。 *输入描述:无 *程序输出:所得结果。 */

1.头文件：btree.h定义数据结构并声明用于完成基本运算的函数。

#ifndef BTREE_H_INCLUDED#define BTREE_H_INCLUDED#define MaxSize 100typedef char ElemType;typedef struct node{    ElemType data;              //数据元素    struct node *lchild;        //指向左孩子    struct node *rchild;        //指向右孩子} BTNode;void CreateBTNode(BTNode *&b,char *str);        //由str串创建二叉链BTNode *FindNode(BTNode *b,ElemType x);     //返回data域为x的节点指针BTNode *LchildNode(BTNode *p);  //返回*p节点的左孩子节点指针BTNode *RchildNode(BTNode *p);  //返回*p节点的右孩子节点指针int BTNodeDepth(BTNode *b); //求二叉树b的深度void DispBTNode(BTNode *b); //以括号表示法输出二叉树void DestroyBTNode(BTNode *&b);  //销毁二叉树#endif // BTREE_H_INCLUDED

2.源文件：btree.cpp实现各个函数

#include <stdio.h>#include <malloc.h>#include "btree.h"void CreateBTNode(BTNode *&b,char *str)     //由str串创建二叉链{    BTNode *St[MaxSize],*p=NULL;    int top=-1,k,j=0;    char ch;    b=NULL;             //建立的二叉树初始时为空    ch=str[j];    while (ch!='\0')    //str未扫描完时循环    {        switch(ch)        {        case '(':            top++;            St[top]=p;            k=1;            break;      //为左节点        case ')':            top--;            break;        case ',':            k=2;            break;                          //为右节点        default:            p=(BTNode *)malloc(sizeof(BTNode));            p->data=ch;            p->lchild=p->rchild=NULL;            if (b==NULL)                    //p指向二叉树的根节点                b=p;            else                            //已建立二叉树根节点            {                switch(k)                {                case 1:                    St[top]->lchild=p;                    break;                case 2:                    St[top]->rchild=p;                    break;                }            }        }        j++;        ch=str[j];    }}BTNode *FindNode(BTNode *b,ElemType x)  //返回data域为x的节点指针{    BTNode *p;    if (b==NULL)        return NULL;    else if (b->data==x)        return b;    else    {        p=FindNode(b->lchild,x);        if (p!=NULL)            return p;        else            return FindNode(b->rchild,x);    }}BTNode *LchildNode(BTNode *p)   //返回*p节点的左孩子节点指针{    return p->lchild;}BTNode *RchildNode(BTNode *p)   //返回*p节点的右孩子节点指针{    return p->rchild;}int BTNodeDepth(BTNode *b)  //求二叉树b的深度{    int lchilddep,rchilddep;    if (b==NULL)        return(0);                          //空树的高度为0    else    {        lchilddep=BTNodeDepth(b->lchild);   //求左子树的高度为lchilddep        rchilddep=BTNodeDepth(b->rchild);   //求右子树的高度为rchilddep        return (lchilddep>rchilddep)? (lchilddep+1):(rchilddep+1);    }}void DispBTNode(BTNode *b)  //以括号表示法输出二叉树{    if (b!=NULL)    {        printf("%c",b->data);        if (b->lchild!=NULL || b->rchild!=NULL)        {            printf("(");            DispBTNode(b->lchild);            if (b->rchild!=NULL) printf(",");            DispBTNode(b->rchild);            printf(")");        }    }}void DestroyBTNode(BTNode *&b)   //销毁二叉树{    if (b!=NULL)    {        DestroyBTNode(b->lchild);        DestroyBTNode(b->rchild);        free(b);    }}

3.测试函数：main.cpp，完成相关测试工作。

（1）计算二叉树节点个数；

<code class="hljs perl has-numbering"><span class="hljs-comment">#include <stdio.h></span><span class="hljs-comment">#include "btree.h"</span><span class="hljs-keyword">int</span> Nodes(BTNode <span class="hljs-variable">*b</span>){    <span class="hljs-keyword">if</span> (b==NULL)        <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;    <span class="hljs-keyword">else</span>        <span class="hljs-keyword">return</span> Nodes(b->lchild)+Nodes(b->rchild)+<span class="hljs-number">1</span>;}<span class="hljs-keyword">int</span> main(){    BTNode <span class="hljs-variable">*b</span>;    CreateBTNode(b,<span class="hljs-string">"A(B(D,E(H(J,K(L,M(,N))))),C(F,G(,I)))"</span>);    <span class="hljs-keyword">printf</span>(<span class="hljs-string">"二叉树节点个数: <span class="hljs-variable">%d</span>\n"</span>, Nodes(b));    DestroyBTNode(b);    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;}</code>

运行结果：

（2）输出所有叶子节点；

<code class="hljs lasso has-numbering"><span class="hljs-variable">#include</span> <span class="hljs-subst"><</span>stdio<span class="hljs-built_in">.</span>h<span class="hljs-subst">></span><span class="hljs-variable">#include</span> <span class="hljs-string">"btree.h"</span><span class="hljs-literal">void</span> DispLeaf(BTNode <span class="hljs-subst">*</span>b){    <span class="hljs-keyword">if</span> (b<span class="hljs-subst">!=</span><span class="hljs-built_in">NULL</span>)    {        <span class="hljs-keyword">if</span> (b<span class="hljs-subst">-></span>lchild<span class="hljs-subst">==</span><span class="hljs-built_in">NULL</span> <span class="hljs-subst">&&</span> b<span class="hljs-subst">-></span>rchild<span class="hljs-subst">==</span><span class="hljs-built_in">NULL</span>)            printf(<span class="hljs-string">"%c "</span>,b<span class="hljs-subst">-></span><span class="hljs-built_in">data</span>);        <span class="hljs-keyword">else</span>        {            DispLeaf(b<span class="hljs-subst">-></span>lchild);            DispLeaf(b<span class="hljs-subst">-></span>rchild);        }    }}int main(){    BTNode <span class="hljs-subst">*</span>b;    CreateBTNode(b,<span class="hljs-string">"A(B(D,E(H(J,K(L,M(,N))))),C(F,G(,I)))"</span>);    printf(<span class="hljs-string">"二叉树中所有的叶子节点是: "</span>);    DispLeaf(b);    printf(<span class="hljs-string">"\n"</span>);    DestroyBTNode(b);    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;}</code>

运行结果:

（3）求二叉树b的叶子节点个数

<code class="hljs lasso has-numbering"><span class="hljs-variable">#include</span> <span class="hljs-subst"><</span>stdio<span class="hljs-built_in">.</span>h<span class="hljs-subst">></span><span class="hljs-variable">#include</span> <span class="hljs-string">"btree.h"</span>int LeafNodes(BTNode <span class="hljs-subst">*</span>b)    <span class="hljs-comment">//求二叉树b的叶子节点个数</span>{    int num1,num2;    <span class="hljs-keyword">if</span> (b<span class="hljs-subst">==</span><span class="hljs-built_in">NULL</span>)        <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;    <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span> (b<span class="hljs-subst">-></span>lchild<span class="hljs-subst">==</span><span class="hljs-built_in">NULL</span> <span class="hljs-subst">&&</span> b<span class="hljs-subst">-></span>rchild<span class="hljs-subst">==</span><span class="hljs-built_in">NULL</span>)        <span class="hljs-keyword">return</span> <span class="hljs-number">1</span>;    <span class="hljs-keyword">else</span>    {        num1<span class="hljs-subst">=</span>LeafNodes(b<span class="hljs-subst">-></span>lchild);        num2<span class="hljs-subst">=</span>LeafNodes(b<span class="hljs-subst">-></span>rchild);        <span class="hljs-keyword">return</span> (num1<span class="hljs-subst">+</span>num2);    }}int main(){    BTNode <span class="hljs-subst">*</span>b;    CreateBTNode(b,<span class="hljs-string">"A(B(D,E(H(J,K(L,M(,N))))),C(F,G(,I)))"</span>);    printf(<span class="hljs-string">"二叉树b的叶子节点个数: %d\n"</span>,LeafNodes(b));    DestroyBTNode(b);    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;}</code>

运行结果：

（4）设计一个算法Level(b,x,h)，返回二叉链b中data值为x的节点的层数。

<code class="hljs perl has-numbering"><span class="hljs-comment">#include <stdio.h></span><span class="hljs-comment">#include "btree.h"</span><span class="hljs-keyword">int</span> Level(BTNode <span class="hljs-variable">*b</span>,ElemType <span class="hljs-keyword">x</span>,<span class="hljs-keyword">int</span> h){    <span class="hljs-keyword">int</span> l;    <span class="hljs-keyword">if</span> (b==NULL)        <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;    <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span> (b->data==<span class="hljs-keyword">x</span>)        <span class="hljs-keyword">return</span> h;    <span class="hljs-keyword">else</span>    {        l=Level(b->lchild,<span class="hljs-keyword">x</span>,h+<span class="hljs-number">1</span>);        <span class="hljs-keyword">if</span> (l==<span class="hljs-number">0</span>)            <span class="hljs-keyword">return</span> Level(b->rchild,<span class="hljs-keyword">x</span>,h+<span class="hljs-number">1</span>);        <span class="hljs-keyword">else</span>            <span class="hljs-keyword">return</span> l;    }}<span class="hljs-keyword">int</span> main(){    BTNode <span class="hljs-variable">*b</span>;    CreateBTNode(b,<span class="hljs-string">"A(B(D,E(H(J,K(L,M(,N))))),C(F,G(,I)))"</span>);    <span class="hljs-keyword">printf</span>(<span class="hljs-string">"值为\'K\'的节点在二叉树中出现在第 <span class="hljs-variable">%d</span> 层上n"</span>,Level(b,<span class="hljs-string">'K'</span>,<span class="hljs-number">1</span>));    DestroyBTNode(b);    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;}</code>

运行结果：

（5）判断二叉树是否相似（关于二叉树t1和t2相似的判断：①t1和t2都是空的二叉树，相似；②t1和t2之一为空，另一不为空，则不相似；③t1的左子树和t2的左子树是相似的，且t1的右子树与t2的右子树是相似的，则t1和t2相似。）

<code class="hljs perl has-numbering"><span class="hljs-comment">#include <stdio.h></span><span class="hljs-comment">#include "btree.h"</span><span class="hljs-keyword">int</span> Like(BTNode <span class="hljs-variable">*b1</span>,BTNode <span class="hljs-variable">*b2</span>){    <span class="hljs-keyword">int</span> like1,like2;    <span class="hljs-keyword">if</span> (b1==NULL && b2==NULL)        <span class="hljs-keyword">return</span> <span class="hljs-number">1</span>;    <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span> (b1==NULL || b2==NULL)        <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;    <span class="hljs-keyword">else</span>    {        like1=Like(b1->lchild,b2->lchild);        like2=Like(b1->rchild,b2->rchild);        <span class="hljs-keyword">return</span> (like1 & like2);    }}<span class="hljs-keyword">int</span> main(){    BTNode <span class="hljs-variable">*b1</span>, <span class="hljs-variable">*b2</span>, <span class="hljs-variable">*b3</span>;    CreateBTNode(b1,<span class="hljs-string">"B(D,E(H(J,K(L,M(,N)))))"</span>);    CreateBTNode(b2,<span class="hljs-string">"A(B(D(,G)),C(E,F))"</span>);    CreateBTNode(b3,<span class="hljs-string">"u(v(w(,x)),y(z,p))"</span>);    <span class="hljs-keyword">if</span>(Like(b1, b2))        <span class="hljs-keyword">printf</span>(<span class="hljs-string">"b1和b2相似\n"</span>);    <span class="hljs-keyword">else</span>        <span class="hljs-keyword">printf</span>(<span class="hljs-string">"b1和b2不相似\n"</span>);    <span class="hljs-keyword">if</span>(Like(b2, b3))        <span class="hljs-keyword">printf</span>(<span class="hljs-string">"b2和b3相似\n"</span>);    <span class="hljs-keyword">else</span>        <span class="hljs-keyword">printf</span>(<span class="hljs-string">"b2和b3不相似\n"</span>);    DestroyBTNode(b1);    DestroyBTNode(b2);    DestroyBTNode(b3);    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;}</code>

运行结果：

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