用二叉树实现存储信息

1、用二叉树实现存储信息 用二叉树实现存储信息.txt老子忽悠孩子叫教 育，孩子忽悠老子叫欺骗，互相忽悠叫代沟。 男人这花花世界，我要用什么颜色来吸引你。#in elude #inelude using n amespaee std;typedef struct _stude ntint id;ehar20 Name;Int Age ；int Math;/（数学成绩）;/学生结构体struct treeitemstruct treeitem *lchild;struct treeitem *rchild;struct _stude nt mystude nt;/二叉树结构体class Stude

2、ntpublic:Stude nt(void);Stude nt(void);void Create(struct treeitem *Tnode );/创建void Find(struct treeitem *tree,int i);/ 查看节点是否存在void Search(Stude nt &tree);查询节点voidChan ge(structtreeitem*Tno de,i nti);/ 修改void DeleteNode(struct treeitem *tree,int i);/删除Stude nt *n ode;/根节点private:bool tj;int i_id,i_

3、age,i_math;char c_ch;.cpp=Stude nt:Stude nt(void)tj=false;Student:Stude nt(void)voidStude nt:Create(structtreeitem*Tnode)/ 创建 1ci nid,c_ch,i_age,i_math;if(id!=null&ch!=&age!=null&math!=null)struct _stude nt *Ts;Ts-id=id;Ts-Name=c_ch;Ts-Age=i_age;Ts-Math=i_math;Tno de=(struct treeitem*)malloc(sizeof(

4、structtreeitem);Tno de-mystude nt=Ts;Create(T no de-lchild); /生成左子树Create(T no de-rchild); /生成右子树coutCreate Complated!e ndl;elseTn ode=NULL; coutvNo Create Complated!id=i)tj=true;return;if(tree-lchild匸NULL)f(tree-lchild-id = i)tj=true;return; if(tree-rchild!=NULL)f(tree-rchild-id = i)tj=true;return;

5、Fin d(tree-lchild,i);Fin d(tree-rchild,i);void Student:Search(Student &tree)if(tree!NULL)if(tree-lchild-lchild=NULL&tree-lc hild-rchild=NULL)retur n tree-lchild;elsef(tree-rchild-lchild=NULL&tree-rchi d-rchild=NULL)return tree-rchild; return Search(tree-lchild); return Search(tree-rchild);treeitemvo

6、idStude nt:Change(struct*Tnode, int i)Find (Tnode, i );if(!tj)coutvv无此节点！ i_age,i_math,c_ch;Tno de-mystude nt.Age=i_age;Tno de-mystude nt.Name=c_ch;Tno de-mystude nt.Math=i_math;void Stude nt:DeleteNode(struct*tree,i nt i)Fin d(tree,i);if(!tj)cout无点!id= i)/的正好是根节点f(no de-lchild = NULL & node-rchildN

7、ULL)/只有一个根节点的情况treeitem要删除delete node;node=NULL;coutvv已删除节点,当前树为空lchild != NULL & node-rchild =NULL)/有左孩子没右孩子treeitem *p = no de;newtreeitem*dtreeitem(Search( no de)-id);d-lchild = tree-lchild;d-rchild = NULL;node=d;delete p;coutID 值 为vvivv的 节点已 rchild != NULL & node-lchild :NULL)/有右孩子没左孩子treeitem *

8、p = no de;treeitem*dtreeitem(Search( no de)-c onten ts);d-lchild= NULL;d-rchild = tree-rchild;被删除newnode =d;tree-lchild = tree-lchild-rchild;delete p;delete p;点已被删除newdelete p;coutID值 为vvivv的 节vve ndl;else/有左、右孩子treeitem *p = no de;treeitem*dtreeitem(Search( no de)-id);d-lchild = tree-lchild;d-child

9、 = tree-rchild;node=d;coutID 值 为vvivv的节点已被删除 lchild!=NULL)左孩子,且正是要查找的if(tree-lchild-id = i)/子是要查找的if(tree-lchild-lchildtree-lchild-rchild != NULL)不是根节点，树还有若根节点的左孩= NULL &treeitem *p = tree-lchild;coutID 值 为vvivv的节点已被删除lchild-lchild tree-lchild-rchild = NULL) delete tree-lchild;tree-lchild = NULL;cou

10、tID值 为vvivv的节点已被删除lchild-lchild!= NULL &tree-lchild-rchild = NULL)treeitem *p = tree-lchild;delete tree-rchild;d-rchild = tree-lchild-rchild;tree-lchild = tree-lchild-lchild;delete p;coutID值 为vvivv的节点已被删除lchild-lchild!= NULL &tree-lchild-rchild != NULL)treeitem *p = tree-lchild;treeitem*d=newtreeite

11、m(Search( no de)-id);d-left = tree-lchildt-lchild;tree-lchild = d;delete p;return;f(tree-rchild匸NULL)/右孩子,且正是要查找的不是根节点，树还有f(tree-rchild-id = i)NULL &f(tree-rchild-lchild tree-rchild-rchild = NULL)tree-rchild = NULL;coutID值 为vvivv的节点已被删除vve ndl;elsef(tree-rchild-lchild!= NULL &tree-rchild-rchild = NU

12、LL)treeitem *p = tree-rchild;tree-rchild = tree-rchild-lchild;delete p;coutID 值为vvivv的节点已被删除 rchild-child= NULL &tree-rchild-rchild != NULL)treeitem *p = tree-rchild;tree-rchild = tree-rchild-rchild;delete p;coutID 值 为vvivv的节点已被删除rchild-lchild tree-rchild-rchild != NULL)treeitem *p = tree-rchild;treeitem*dnew