About half of large data structures
The set of operations binary search tree
This problem required to achieve a given species to five binary search tree operation.
Function interface definition:
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 );
Wherein the BinTree
structure is defined as follows:
typedef struct TNode *Position;
typedef Position BinTree;
struct TNode{
ElementType Data;
BinTree Left;
BinTree Right;
};
- Function
Insert
TheX
insertion binary search treeBST
and returns a pointer to the root of the result of the tree; - Function
Delete
toX
the binary search treeBST
deletion and returns the pointer to the root of the result tree; ifX
not in the tree, then print a lineNot Found
root pointer and return the original tree; - Function
Find
in the binary search treeBST
found inX
, it returns a pointer to the node; if no null pointer is returned; - Function
FindMin
returns the binary search treeBST
pointer smallest element nodes; - Function
FindMax
returns the binary search treeBST
pointer largest element nodes. - Referee test program example :
-
#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 */
Sample input:
-
10 5 8 6 2 4 1 0 10 9 7 5 6 3 10 0 5 5 5 7 0 10 3
Sample output:
-
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
Subject to answer:
-
BinTree Insert(BinTree BST, ElementType X) { if (BST == NULL) { BST = (BinTree)malloc(sizeof(BinTree)); BST->Data = X; BST->Left = NULL; BST->Right = NULL; } if (X > BST->Data) BST->Right = Insert(BST->Right, X); if (X < BST->Data) BST->Left = Insert(BST->Left, X); return BST; } // 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; BinTree Delete(BinTree BST, ElementType X) { if (BST == NULL) { printf("Not Found\n"); return NULL; } if (X > BST->Data) { BST->Right = Delete(BST->Right, X); } else if (X < BST->Data) { BST->Left = Delete(BST->Left, X); } else if (X == BST->Data) { Position node; if (BST->Left && BST->Right) { node = FindMin(BST->Right); BST->Data = node->Data; BST->Right = Delete(BST->Right, BST->Data); } else { node = BST; if (BST->Left == NULL) BST = BST->Right; else if (BST->Right == NULL) BST = BST->Left; free(node); } } return BST; } //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; Position Find(BinTree BST, ElementType X) { while (BST) { if (X == BST->Data) return BST; else if (X < BST->Data) BST = BST->Left; else if (X > BST->Data) BST = BST->Right; } return NULL; } //函数FindMin返回二叉搜索树BST中最小元结点的指针; Position FindMin(BinTree BST) { if (BST == NULL) return NULL; while (BST->Left) BST = BST->Left; return BST; } //函数FindMax返回二叉搜索树BST中最大元结点的指针。 Position FindMax(BinTree BST) { if (BST == NULL) return NULL; while (BST->Right) BST = BST->Right; return BST; }