题目
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
运算结果
Case | Hint | Result | Run Time | Memory |
---|---|---|---|---|
0 | sample等价 | Accepted | 2 ms | 296 KB |
1 | 左斜,Find小于最小、大于最大,delete小于最小、大于最大、头尾和中间 | Accepted | 2 ms | 228 KB |
2 | 右斜,Find小于最小、大于最大,delete小于最小、大于最大、头尾和中间 | Accepted | 2 ms | 384 KB |
3 | 只有1个结点,删除成空树 | Accepted | 2 ms | 228 KB |
4 | 空树 | Accepted | 2 ms | 296 KB |
程序
#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 */
//将X插入二叉搜索树BST并返回结果树的根结点指针
BinTree Insert( BinTree BST, ElementType X );
//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
BinTree Delete( BinTree BST, ElementType X );
//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
Position Find( BinTree BST, ElementType X );
//返回二叉搜索树BST中最小元结点的指针
Position FindMin( BinTree BST );
//返回二叉搜索树BST中最大元结点的指针
Position FindMax( BinTree BST );
int main()
{
BinTree BST, MinP, MaxP, Tmp;
ElementType X;
int N, i;
BST = NULL;
/*//以下为正式算法
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");
*/
//测试用例
N = 10;
int arr[10] = {5,8,6,2,4,1,0,10,9,7};
for ( i=0; i<N; i++ ) {
X = arr[i];
BST = Insert(BST, X);
}
printf("Preorder:"); PreorderTraversal(BST); printf("\n");
MinP = FindMin(BST);
MaxP = FindMax(BST);
N = 5;
int arr1[5] = {6,3,10,0,5};
for( i=0; i<N; i++ ) {
X = arr1[i];
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);
}
}
N = 5;
int arr2[5] = {5,7,0,10,3};
for( i=0; i<N; i++ ) {
X = arr2[i];
BST = Delete(BST, X);
}
printf("Inorder:"); InorderTraversal(BST); printf("\n");
//以上为测试用例
return 0;
}
//以下为测试函数
void PreorderTraversal( BinTree BT )
{
if(BT){
printf("%d ", BT->Data);
PreorderTraversal(BT->Left);
PreorderTraversal(BT->Right);
}
}
void InorderTraversal( BinTree BT )
{
if(BT){
InorderTraversal(BT->Left);
printf("%d ", BT->Data);
InorderTraversal(BT->Right);
}
}
//以上为测试函数
/* 你的代码将被嵌在这里 */
BinTree Insert( BinTree BST, ElementType X )
{
//递归出口设计:遍历到空结点
if(!BST){
BST = (BinTree)malloc(sizeof(struct TNode));
BST->Data = X;
BST->Left = BST->Right = NULL;
return BST;
}
//递归逻辑:
//1、判断X归属于哪个子树,并插入
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 )
{
//递归出口条件:1、未找到,则打印Not Found,并返回原树的根节点指针
//2、找到,删除结点,并返回结果树的根结点指针
if(!BST){
//printf("%d ", X);
printf("Not Found");printf("\n");
return NULL;
}
else if(X == BST->Data){
//删除的是叶结点:找到该结点,并且左右孩子为空
//删除方法:直接删除,并返回空指针
if((BST->Left == NULL) && (BST->Right == NULL)){
free(BST);
return NULL;
}
//删除的结点只有一个孩子:只有左孩子或者右孩子
//删除方法:返回孩子结点
if((BST->Left != NULL) && (BST->Right == NULL)){
BinTree LeftSubTree;
LeftSubTree = BST->Left;
free(BST);
return LeftSubTree;
}
if((BST->Left == NULL) && (BST->Right != NULL)){
BinTree RightSubTree;
RightSubTree = BST->Right;
free(BST);
return RightSubTree;
}
//删除的结点有两个孩子
//删除方法:用右子树的最小元素或左子树的最大元素代替此结点(此处选前者)
if((BST->Left != NULL) && (BST->Right != NULL)){
BinTree MinRightSubNode = FindMin(BST->Right);
BST->Data = MinRightSubNode->Data;
BST->Right = Delete(BST->Right, BST->Data);
return BST;
}
}
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;
}
}
Position Find( BinTree BST, ElementType X )
{
//递归出口条件:1、未找到,即指针为空,返回空指针;2、找到,返回指针
if(!BST){
return NULL;
}
else if(X == BST->Data){
return BST;
}
else if(X < BST->Data){
Find(BST->Left, X);
}
else if(X > BST->Data){
Find(BST->Right, X);
}
}
Position FindMin( BinTree BST )
{
if(BST){
if(!BST->Left){
return BST;
}
FindMin(BST->Left);
}
else
return BST;
}
Position FindMax( BinTree BST )
{
if(BST){
if(!BST->Right){
return BST;
}
FindMax(BST->Right);
}
else
return BST;
}