第十週上機實踐—項目3—利用二叉遍歷思想解決問題

/*
 *Copyright(c) 2015,煙臺大學計算機學院
 *All rights reserved.
 *文件名稱:test.cpp
 *作者:林莉
 *完成日期:2015年9月11日
 *版本:v1.0
 *
 *問題描述:假設二叉樹採用二叉鏈存儲結構存儲，分別實現以下算法，並在程序中完成測試。
 *輸入描述:無
 *程序輸出:所得結果。
 */

1.頭文件：btree.h定義數據結構並聲明用於完成基本運算的函數。

#ifndef BTREE_H_INCLUDED
#define BTREE_H_INCLUDED

#define MaxSize 100
typedef char ElemType;
typedef struct node
{
    ElemType data;              //數據元素
    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串創建二叉鏈
BTNode *FindNode(BTNode *b,ElemType x);     //返回data域爲x的節點指針
BTNode *LchildNode(BTNode *p);  //返回*p節點的左孩子節點指針
BTNode *RchildNode(BTNode *p);  //返回*p節點的右孩子節點指針
int BTNodeDepth(BTNode *b); //求二叉樹b的深度
void DispBTNode(BTNode *b); //以括號表示法輸出二叉樹
void DestroyBTNode(BTNode *&b);  //銷燬二叉樹

#endif // BTREE_H_INCLUDED

2.源文件：btree.cpp實現各個函數

#include <stdio.h>
#include <malloc.h>
#include "btree.h"

void CreateBTNode(BTNode *&b,char *str)     //由str串創建二叉鏈
{
    BTNode *St[MaxSize],*p=NULL;
    int top=-1,k,j=0;
    char ch;
    b=NULL;             //建立的二叉樹初始時爲空
    ch=str[j];
    while (ch!='\0')    //str未掃描完時循環
    {
        switch(ch)
        {
        case '(':
            top++;
            St[top]=p;
            k=1;
            break;      //爲左節點
        case ')':
            top--;
            break;
        case ',':
            k=2;
            break;                          //爲右節點
        default:
            p=(BTNode *)malloc(sizeof(BTNode));
            p->data=ch;
            p->lchild=p->rchild=NULL;
            if (b==NULL)                    //p指向二叉樹的根節點
                b=p;
            else                            //已建立二叉樹根節點
            {
                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域爲x的節點指針
{
    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節點的左孩子節點指針
{
    return p->lchild;
}
BTNode *RchildNode(BTNode *p)   //返回*p節點的右孩子節點指針
{
    return p->rchild;
}
int BTNodeDepth(BTNode *b)  //求二叉樹b的深度
{
    int lchilddep,rchilddep;
    if (b==NULL)
        return(0);                          //空樹的高度爲0
    else
    {
        lchilddep=BTNodeDepth(b->lchild);   //求左子樹的高度爲lchilddep
        rchilddep=BTNodeDepth(b->rchild);   //求右子樹的高度爲rchilddep
        return (lchilddep>rchilddep)? (lchilddep+1):(rchilddep+1);
    }
}
void DispBTNode(BTNode *b)  //以括號表示法輸出二叉樹
{
    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)   //銷燬二叉樹
{
    if (b!=NULL)
    {
        DestroyBTNode(b->lchild);
        DestroyBTNode(b->rchild);
        free(b);
    }
}

3.測試函數：main.cpp，完成相關測試工作。

（1）計算二叉樹節點個數；

<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">"二叉樹節點個數: <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>

運行結果：

（2）輸出所有葉子節點；

<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">"二叉樹中所有的葉子節點是: "</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>

運行結果:

（3）求二叉樹b的葉子節點個數

<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">//求二叉樹b的葉子節點個數</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">"二叉樹b的葉子節點個數: %d\n"</span>,LeafNodes(b));
    DestroyBTNode(b);
    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;
}</code>

運行結果：

（4）設計一個算法Level(b,x,h)，返回二叉鏈b中data值爲x的節點的層數。

<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">"值爲\'K\'的節點在二叉樹中出現在第 <span class="hljs-variable">%d</span> 層上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>

 

運行結果：

（5）判斷二叉樹是否相似（關於二叉樹t1和t2相似的判斷：①t1和t2都是空的二叉樹，相似；②t1和t2之一爲空，另一不爲空，則不相似；③t1的左子樹和t2的左子樹是相似的，且t1的右子樹與t2的右子樹是相似的，則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>

運行結果：