04-树7 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
题解:
BinTree Insert( BinTree BST, ElementType X ) { if(BST == NULL) { BST = (BinTree)malloc(sizeof(struct TNode)); BST->Data = X; BST->Left = NULL; BST->Right = NULL; } else { if(BST->Data > X) { BST->Left = Insert(BST->Left, X); } else { BST->Right = Insert(BST->Right, X); } } return BST; } BinTree Delete( BinTree BST, ElementType X ){ Position Tmp; if(!BST) printf("Not Found\n"); else { if( X < BST->Data) BST ->Left = Delete(BST->Left, X); /* 左子树递归删除 */ else if(X > BST->Data ) BST ->Right = Delete(BST->Right , X); /* 右子树递归删除*/ else { /* 找到需要删除的结点 */ if(BST->Left && BST->Right) { /* 被删除的结点有左右子结点 */ Tmp=FindMin(BST->Right); /* 在右子树中找到最小结点填充删除结点 */ BST->Data = Tmp ->Data; BST->Right=Delete(BST->Right,BST->Data);/* 递归删除要删除结点的右子树中最小元素 */ }else { /* 被删除结点有一个或没有子结点*/ Tmp = BST; if(!BST->Left) BST = BST->Right; /*有右孩子或者没孩子*/ else if(!BST->Right) BST = BST->Left;/*有左孩子,一定要加else,不然BST可能是NULL,会段错误*/ free(Tmp); /*如无左右孩子直接删除*/ } } } return BST; } Position Find( BinTree BST, ElementType X ) { Position F = NULL; if(BST) { if(BST->Data == X) { F = BST; } else { if(BST->Data > X) { F = Find(BST->Left, X); } else { F = Find(BST->Right, X); } } } else { return NULL; } return F; } Position FindMin( BinTree BST ) { if (BST) { Position Min = BST; while(Min->Left) Min = Min->Left; return Min; } else { return NULL; } } Position FindMax( BinTree BST ) { if (BST) { Position Max = BST; while(Max->Right) Max = Max->Right; return Max; } else { return NULL; } }
//二叉搜索树的前,中序遍历;
void PreorderTraversal( BinTree BT ) { if(BT) { printf("%d ",BT->Data); PreorderTraversal( BT->Left ); PreorderTraversal( BT->Right ); } else { return; } } /* 先序遍历,由裁判实现,细节不表 */ void InorderTraversal( BinTree BT ) { if(BT) { InorderTraversal( BT->Left ); printf("%d ",BT->Data); InorderTraversal( BT->Right ); } else { return; } } /* 中序遍历,由裁判实现,细节不表 */