6-12 二叉搜索树的操作集 (30分)

本题要求实现给定二叉搜索树的5种常用操作。

函数接口定义:
BinTree Insert( BinTree BST, ElementType X );
BinTree Delete( BinTree BST, ElementType X );
Position Find( BinTree BST, ElementType X );
Position FindMin( BinTree BST );
Position FindMax( BinTree BST );
其中BinTree结构定义如下:

typedef struct TNode *Position;
typedef Position BinTree;
struct TNode{
    
    
    ElementType Data;
    BinTree Left;
    BinTree Right;
};

函数Insert将X插入二叉搜索树BST并返回结果树的根结点指针;
函数Delete将X从二叉搜索树BST中删除,并返回结果树的根结点指针;如果X不在树中,则打印一行Not Found并返回原树的根结点指针;
函数Find在二叉搜索树BST中找到X,返回该结点的指针;如果找不到则返回空指针;
函数FindMin返回二叉搜索树BST中最小元结点的指针;
函数FindMax返回二叉搜索树BST中最大元结点的指针。
裁判测试程序样例:

#include <stdio.h>
#include <stdlib.h>

typedef int ElementType;
typedef struct TNode *Position;
typedef Position BinTree;
struct TNode{
    
    
    ElementType Data;
    BinTree Left;
    BinTree Right;
};

void PreorderTraversal( BinTree BT ); /* 先序遍历,由裁判实现,细节不表 */
void InorderTraversal( BinTree BT );  /* 中序遍历,由裁判实现,细节不表 */

BinTree Insert( BinTree BST, ElementType X );
BinTree Delete( BinTree BST, ElementType X );
Position Find( BinTree BST, ElementType X );
Position FindMin( BinTree BST );
Position FindMax( BinTree BST );

int main()
{
    
    
    BinTree BST, MinP, MaxP, Tmp;
    ElementType X;
    int N, i;

    BST = NULL;
    scanf("%d", &N);
    for ( i=0; i<N; i++ ) {
    
    
        scanf("%d", &X);
        BST = Insert(BST, X);
    }
    printf("Preorder:"); PreorderTraversal(BST); printf("\n");
    MinP = FindMin(BST);
    MaxP = FindMax(BST);
    scanf("%d", &N);
    for( i=0; i<N; i++ ) {
    
    
        scanf("%d", &X);
        Tmp = Find(BST, X);
        if (Tmp == NULL) printf("%d is not found\n", X);
        else {
    
    
            printf("%d is found\n", Tmp->Data);
            if (Tmp==MinP) printf("%d is the smallest key\n", Tmp->Data);
            if (Tmp==MaxP) printf("%d is the largest key\n", Tmp->Data);
        }
    }
    scanf("%d", &N);
    for( i=0; i<N; i++ ) {
    
    
        scanf("%d", &X);
        BST = Delete(BST, X);
    }
    printf("Inorder:"); InorderTraversal(BST); printf("\n");

    return 0;
}
/* 你的代码将被嵌在这里 */

输入样例:
10
5 8 6 2 4 1 0 10 9 7
5
6 3 10 0 5
5
5 7 0 10 3
输出样例:
Preorder: 5 2 1 0 4 8 6 7 10 9
6 is found
3 is not found
10 is found
10 is the largest key
0 is found
0 is the smallest key
5 is found
Not Found
Inorder: 1 2 4 6 8 9

Position Find(BinTree BST,ElementType X){
    
    
    if(BST==NULL)return NULL;
    else if(X<BST->Data)return Find(BST->Left,X);
    else if(X>BST->Data)return Find(BST->Right,X);
    else return BST;
    
}
Position FindMin(BinTree BST){
    
    
    if(BST==NULL)return NULL;
    else if(BST->Left==NULL)return BST;
    else return FindMin(BST->Left);
}
Position FindMax(BinTree BST){
    
    
    if(BST==NULL)return NULL;
    else if(BST->Right==NULL)return BST;
    else return FindMax(BST->Right);
}
BinTree Insert(BinTree BST,ElementType X){
    
    
    if(BST==NULL){
    
    
        BST=(BinTree)malloc(sizeof(struct TNode));
        BST->Data=X;
        BST->Left=BST->Right=NULL;
    }
    else if(X<BST->Data)BST->Left=Insert(BST->Left,X);
    else if(X>BST->Data)BST->Right=Insert(BST->Right,X);
    return BST;
}
BinTree Delete(BinTree BST,ElementType X){
    
    
    if(BST==NULL){
    
    
        printf("Not Found\n");
    }
    else{
    
    
        if(X<BST->Data)BST->Left=Delete(BST->Left,X);
        else if(X>BST->Data)BST->Right=Delete(BST->Right,X);
        else if(X==BST->Data){
    
    
            if(BST->Left&&BST->Right){
    
    //若左右儿子都存在,寻找右儿子的最小值替换
                BinTree pos=FindMin(BST->Right);
                BST->Data=pos->Data;
                BST->Right=Delete(BST->Right,BST->Data);
            }
            else{
    
    //只有一个儿子
                if(BST->Left==NULL)BST=BST->Right;
                else if(BST->Right==NULL)BST=BST->Left;
            }
        }
    }
    return BST;
}

只有一个儿子的情况在这里插入图片描述
两个儿子都存在的情况
在这里插入图片描述

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转载自blog.csdn.net/qq_40924271/article/details/109171311