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
通过代码:
/* 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 ){
if( !BST ){
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;
}
/* 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 ){
Position Tmp;
if( !BST )
printf("Not Found\n"); //记得\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 ){
Tmp = FindMin( BST->Right );
BST->Data = Tmp->Data;
BST->Right = Delete(BST->Right, BST->Data);
}else{
Tmp = BST;
if( !BST->Left )
BST = BST->Right;
else if( !BST->Right )
BST = BST->Left;
free( Tmp );
}
}
return BST;
}
/* 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 ){
while( BST ){
if( X > BST->Data )
BST = BST->Right; //X比这个节点的数值大,往右子树找
else if( X < BST->Data )
BST = BST->Left; //X比这个节点的数值小,往左子树找
else
return BST; //等于X——找到啦
}
return NULL; //查找失败
}
/* 返回二叉搜索树BST中最小元结点的指针(递归写法) */
Position FindMin( BinTree BST ){
if( !BST ) //是空树
return NULL;
else if( !BST->Left )
return BST; //找到最左叶节点(最左叶节点就是最小值)
else
return FindMin( BST->Left ); //继续往左分支查找
}
/* 返回二叉搜索树BST中最大元结点的指针(非递归写法) */
Position FindMax( BinTree BST ){
if( !BST ) //是空树
return NULL;
else {
while( BST->Right ){ //一直往右节点查找,直到遇到NULL(此处遇到NULL时不赋值给BST)
BST = BST->Right;
}
}
return BST; //继续往右分支查找
}