二叉树

此次作业中公共数据类型定义在如下同一个头文件中：//bTree.h : 包含二叉树常用的存储结构和二叉树的常用算法#include <stdio.h>#define MaxNode 11//s树or二叉树的最大节点个数typedef char DataType;//定义节点数据类型typedef struct seqnode{//定义顺序储存结构的树的节点DataType data;//节点数据元素int parent;//指向父亲的数组伪指针}seqNode;typedef seqNode SqTree[MaxNode];//定义顺序储存结构的树typedef DataType SqBTree[MaxNode];//定义顺序存储结构的二叉树typedef struct chnnode{//定义链式存储结构的二叉树节点DataType data;//节点数据元素struct chnnode * lchild;//左孩子指针struct chnnode * rchild;//右孩子指针}chnNode;typedef struct chnhead{//定义链式存储节点二叉树的头结点int nodeCount;//记录节点个数struct chnnode * root;//指向根节点}chnHead;typedef chnHead CnBTree;//定义链式存储结构的二叉树7.4:// problem_7.4.cpp : 定义控制台应用程序的入口点。////probem:将n个节点以顺序存放的完全二叉树转化为二叉链存储结构#include "stdafx.h"#include "bTree.h"chnNode * CnBTreeCreate(DataType * sbt,int n){//递归方法创建完全二叉树,对于错误的完全二叉树，则只会在错误之前的节点正常chnNode *temp = NULL;if (n == 0)temp=CnBTreeCreate(sbt,n+1);else{if (*(sbt + n) == '#')temp = NULL;else{temp = (chnNode *)malloc(sizeof(chnNode));temp->data = *(sbt + n);if (2 * n < MaxNode)temp->lchild = CnBTreeCreate(sbt, 2 * n);elsetemp->lchild = NULL;if (2 * n + 1 < MaxNode)temp->rchild = CnBTreeCreate(sbt, 2 * n + 1);elsetemp->rchild = NULL;}}return temp;}int _tmain(int argc, _TCHAR* argv[]){SqBTree Sbt = "#ABC#EFGHI";CnBTree Cnt;Cnt.nodeCount = strlen(Sbt) - 1;Cnt.root=CnBTreeCreate(Sbt,0);system("PAUSE");return 0;}// problem_7.5.cpp : 定义控制台应用程序的入口点。void CnBTree2SeqBTree(chnNode * chn,DataType *sbt,int n){if (n == 0){*sbt = '#';CnBTree2SeqBTree(chn, sbt, n + 1);}else{if (chn == NULL){*(sbt + n) = '#';return;}else{*(sbt + n) = chn->data;if (2 * n < MaxNode - 1)CnBTree2SeqBTree(chn->lchild, sbt, 2 * n);if (2 * n + 1 < MaxNode - 1)CnBTree2SeqBTree(chn->rchild, sbt, 2 * n + 1);}}}7.6// problem_7.6.cpp : 定义控制台应用程序的入口点。//bool IsCompleteBTree(CnBTree * cnt){SqBTree St = "##########";CnBTree2SeqBTree(cnt->root,St,0);for (int i = 1; i < MaxNode; i++)if (*(St + i) == '#')return false;return true;}7.7．bTree.h新增加两种数据结构typedef struct{chnNode * cnNode[MaxNode];int top;}BTreeStack;typedef struct{struct{chnNode * cnNode;int k;}WayNode[MaxNode];int top;}WayStack;void CreateChnBTree(chnNode * &cn , char * str){BTreeStack bts;bts.top = -1;int i = 0,k=0;char ch;chnNode * temp = NULL;while ((ch=*(str + i)) != '\0'){switch (ch){ case '(': bts.cnNode[++bts.top] = temp; k = 1; break; case ',': k = 2; break; case ')': k = 1; bts.top--; break; default: temp = (chnNode *)malloc(sizeof(chnNode)); temp->data = ch; temp->lchild = NULL; temp->rchild = NULL; switch (k){ case 0: cn = temp; break; case 1: bts.cnNode[bts.top]->lchild = temp; break; case 2: bts.cnNode[bts.top]->rchild = temp; break; } break;}i++;}}bool FindWaybyStack(WayStack *ws, chnNode * cn, char e){ws->WayNode[++ws->top].cnNode = cn;ws->WayNode[ws->top].k = 0;chnNode * temp = NULL;while (ws->top != -1){if (ws->WayNode[ws->top].cnNode->data == e)//说明找到一条路径return true;else{if (ws->WayNode[ws->top].k == 0){ws->WayNode[ws->top].k++;temp = ws->WayNode[ws->top].cnNode->lchild;if (temp == NULL)continue;ws->WayNode[++ws->top].cnNode = temp;ws->WayNode[ws->top].k = 0;}else if (ws->WayNode[ws->top].k == 1){ ws->WayNode[ws->top].k++;temp = ws->WayNode[ws->top].cnNode->rchild; if (temp == NULL)continue;ws->WayNode[++ws->top].cnNode = temp;ws->WayNode[ws->top].k = 0;}elsews->top--;}}return false;}int _tmain(int argc, _TCHAR* argv[]){char * btStr = "A(B(D(G),E(,H)),C(,F(,I(J))))";CnBTree cbt;cbt.root = NULL;CreateChnBTree(cbt.root,btStr);WayStack wSt;wSt.top = -1;bool mark=FindWaybyStack(&wSt,cbt.root,'I');if (mark == true){printf("路径如下:\n");for (int i = 0; i <= wSt.top; i++)printf("->%c", wSt.WayNode[i].cnNode->data);}elseprintf("该元素不存在！\n");system("PAUSE");return 0;}7.8：chnNode * SwapNode(chnNode * cn){chnNode * temp;if (cn == NULL)return NULL;else{temp = (chnNode *)malloc(sizeof(chnNode));temp->data = cn->data;temp->lchild = SwapNode(cn->rchild);temp->rchild = SwapNode(cn->lchild);return temp;}}7.9：利用7.8的交换函数，直接将两个左子树进行比较即可bool IsSmmetry(chnNode *cn1,chnNode *cn2){if ((cn1 == NULL) ^ (cn2 == NULL))return false;else{if (cn1 != NULL&&cn2 != NULL)return IsSmmetry(cn1->lchild, cn2->lchild) & IsSmmetry(cn1->rchild, cn2->rchild);elsereturn true;}}7.10：bool IsUnionAncestor(char e,char e1,char e2,chnNode *cn){WayStack ws1, ws2;bool mark1 = false, mark2 = false;ws1.top = ws2.top = -1;if (FindWaybyStack(&ws1, cn, e1) && FindWaybyStack(&ws2, cn, e2)){for (int i = 0; i <= ws1.top; i++)if (ws1.WayNode[i].cnNode->data == e){mark1 = true;break;}for (int i = 0; i <= ws2.top; i++)if (ws2.WayNode[i].cnNode->data == e){mark2 = true;break;}return mark1 && mark2;}elsereturn false;}

一点点作业，记录一下