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
思路：二叉搜索树忘了??，看完之后曾国强不想用递归。
BinTree Insert(BinTree BST, ElementType X) { if (BST == NULL){ BST = (BinTree)malloc(sizeof(struct TNode)); BST->Data = X; BST->Left = NULL; BST->Right = NULL; return BST; } BinTree bin = BST; while (bin){ if (X < bin->Data){ if (bin->Left == NULL){ bin->Left = (BinTree)malloc(sizeof(struct TNode)); bin->Left->Data = X; bin->Left->Left = NULL; bin->Left->Right = NULL; break; } else bin = bin->Left; } else if (bin->Data < X) { if (bin->Right == NULL){ bin->Right = (BinTree)malloc(sizeof(struct TNode)); bin->Right->Data = X; bin->Right->Left = NULL; bin->Right->Right = NULL; break; } else bin = bin->Right; } } return BST; } BinTree Delete(BinTree BST, ElementType X) { if (!BST) 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 (BST->Left&&BST->Right){ Position pos = FindMin(BST->Right); BST->Data = pos->Data; BST->Right = Delete(BST->Right, BST->Data); } else if (!BST->Left){ Position pos = BST; BST = BST->Right; free(pos); } else if (!BST->Right){ Position pos = BST; BST = BST->Left; free(pos); } } } return BST; } Position Find(BinTree BST, ElementType X) { if (!BST)return BST; Position pos = BST; while (pos){ if (pos->Data == X)return pos; if (pos->Data > X){ if (pos->Left == NULL) return pos->Left; else pos = pos->Left; } if (pos->Data < X){ if (pos->Right == NULL) return pos->Right; else pos = pos->Right; } } } Position FindMin(BinTree BST) { if (!BST)return BST; Position pos = BST; while (pos->Left) pos = pos->Left; return pos; } Position FindMax(BinTree BST) { if (!BST)return BST; Position pos = BST; while (pos->Right) pos = pos->Right; return pos; }
