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

代码：

Position Find(BinTree BST, ElementType X){
	while (BST){
		if (X > BST->Data) BST = BST->Right;
		else if (X < BST->Data) BST = BST->Left;//此处一定要加else，下同
		else if(X==BST->Data) return BST;
	}
	return NULL;
}

Position FindMin( BinTree BST ){
    if(!BST) return NULL;
    while(BST->Left) BST = BST->Left;
    return BST;
}

Position FindMax( BinTree BST ){
    if(!BST) return NULL;
    while(BST->Right) BST = BST->Right;
    return BST;
}

BinTree Insert( BinTree BST, ElementType X ){
    if(!BST){
        BST = (BinTree)malloc(sizeof(struct TNode));
        BST->Data = X;
        BST->Left = NULL;
        BST->Right = NULL;
    }
    if(X < BST->Data) BST->Left = Insert(BST->Left, X);
    if(X > BST->Data) BST->Right = Insert(BST->Right, X);
    return BST;
}

BinTree Delete( BinTree BST, ElementType X ){
    Position temp;
    if(!BST){
        printf("Not Found\n");
        return BST;
    }
    if(X < BST->Data) BST->Left = Delete(BST->Left, X);
    if(X > BST->Data) BST->Right = Delete(BST->Right, X);
    if(X==BST->Data){
        if(BST->Left&&BST->Right){
            temp = FindMin(BST->Right);
            BST->Data = temp->Data;
            BST->Right = Delete(BST->Right, BST->Data);
        }
        else{
            if(BST->Left) BST = BST->Left;
            else BST = BST->Right;
        }
    }
    return BST;
}

总结：

1. else表示与之前的if互斥，Find()函数中，第一个if后改变了BST值，若之后不加else则会不断循环。