基于二叉搜索树的符号表和BST排序

来源：互联网发布：linux挂载ntfs 只读编辑：程序博客网时间：2024/06/05 13:21
原代码如下：
#include <stdlib.h>#include <stdio.h>//#define Key inttypedef int Key;struct Item{Key key;char c;};typedef struct STnode* link;struct STnode{Item item ; link l,r; int N;};static link head , z ;static struct Item NULLitem ;Key key(Item item){return item.key;}//创建一个节点 link NEW(Item item, link l,link r, int N){link x = (link)malloc(sizeof *x);x->item = item;x->l = l;x->r = r;x->N = N;if(head==z)head=x; //这句话不能少！！！ return x;}//初始化 void STinit(){head = ( z = NEW(NULLitem,0,0,0));}//节点个数 int STcount(){return head->N; } //搜索子程序 Item searchR(link h, Key v){Key t = key(h->item);if(h==z)return NULLitem;if(v==t) return h->item;if(v<t) return searchR(h->l,v);else return searchR(h->r,v);}//搜索主程序 Item STsearch(Key v){return searchR(head,v);}//插入子程序 link insertR(link h, Item item){Key v = key(item), t = key(h->item);if(h==z)return NEW(item,z,z,1);if(v<t) h->l = insertR(h->l,item);else h->r = insertR(h->r,item);(h->N)++;return h;}//插入主程序 link STinsert(Item item){return insertR(head,item);}//删除子程序 Item deleteR(link F){Item tmp; link p;if(F->l==NULL){p = F;tmp = F->item;F = F->r;free(p);return tmp;}else return deleteR(F->l);}//删除子程序 void deleteRR(link h , Key v){if(h!=NULL){Key t = key(h->item);if(v<t) deleteRR(h->l,v);else if(v>t) deleteRR(h->r,v);elseif(h->l==NULL) { //处理只有一颗子树或没有子树的情况  1 link p = h->r; h=p; free(p);}else if(h->r==NULL){ //处理只有一颗子树或没有子树的情况  2 link p = h->l; h=p; free(p);} else h->item= deleteR(h->r); //如果待删除的节点既有左子树又有右子树//则用该节点右子树的最左下节点替换之，维持二叉搜索树 }} //删除主程序 void STdelete(Key v){ deleteRR(head,v);}void sortR(link h){if(h==z)return;sortR(h->l);if(h->item.key!=0)printf("%d ",h->item.key);sortR(h->r);}void STsort(){sortR(head);}void test(){struct Item item1 = {322,'a'};struct Item item2 = {23,'a'};struct Item item3 = {2,'a'};struct Item item4 = {332,'a'};STinit();STinsert(item1);STinsert(item2);STinsert(item4);STinsert(item3);STsort();printf("\n");struct Item item11 = STsearch(23); printf("%d\n",item11.key);//STdelete(23);STdelete(322);STsort();}main(){test();}
运行结果
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