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：
代码语言：javascript复制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：
代码语言：javascript复制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：
代码语言：javascript复制#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 foundn", X);
else {
printf("%d is foundn", Tmp->Data);
if (Tmp==MinP) printf("%d is the smallest keyn", Tmp->Data);
if (Tmp==MaxP) printf("%d is the largest keyn", 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：
代码语言：javascript复制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:
代码语言：javascript复制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代码实现（gcc 6.5.0）
代码语言：javascript复制BinTree Delete( BinTree BST, ElementType X ){
if(BST == NULL){ printf("Not Foundn"); return BST;}
else if(BST->Data == X){
if(BST->Left != NULL && BST->Right != NULL){
BinTree p = FindMax(BST->Left);
//while(p->Right) p = p->Right;
p->Right = BST->Right;
p = BST->Left;
free(BST);
return p;
}
else if(BST->Left != NULL && BST->Right == NULL){
BinTree p = BST->Left;
free(BST);
return p;
}
else if(BST->Right != NULL && BST->Left == NULL){
BinTree p = BST->Right;
free(BST);
return p;
}
else if((BST->Left != NULL) && (BST->Right == NULL)){
BinTree p = BST;
free(p);
return NULL;
}
}
else if(X < BST->Data) {
BST->Left = Delete(BST->Left, X);
return BST;
}
else if(X > BST->Data){ BST->Right = Delete(BST->Right, X); return BST;}
}
BinTree Insert( BinTree BST, ElementType X ) {
if(BST == NULL){
BST = (BinTree)malloc(sizeof(Position));
BST->Data = X;
BST->Left = NULL;
BST->Right = NULL;
}
else if(X > BST->Data) BST->Right =Insert(BST->Right, X);
else if(X < BST->Data) BST->Left = Insert(BST->Left, X);
return BST;
}//diyidabug
Position Find( BinTree BST, ElementType X ){
if(BST == NULL) return 0;
else if(BST->Data == X) return BST;
else if(X < BST->Data) Find(BST->Left, X);
else Find(BST->Right, X);
}
Position FindMin( BinTree BST ){
if(!BST) return BST;
if(BST->Left == NULL) return BST;
else FindMin(BST->Left);
}
Position FindMax( BinTree BST ){
if(!BST) return BST;
if(BST->Right == NULL) return BST;
else FindMax(BST->Right);
}
