考研复习（7）树的基本操作

这次是树的基本操作，包括生成树，求树深度，求叶子节点个数。。。
附上二叉树的几个重要性质及证明（考研必考）。
1.在二叉树的第i层至多有2^（i-1）个结点；
2.深度为k的二叉树至多有2^(k)-1 个结点；
3.对任何一棵二叉树T，如果其终端结点数为n0，度为2的结点数为n2，则n0=n2+1；（n=n0+n1+n2=n1+2n2+1）;
4.具有n哥节点的完全二叉树深度为【log2n】+1；

#include <stdio.h>#include<stdlib.h>#define maxSize 20#define maxWidth 20typedef struct node{char data;struct node *left,*right;}Btree;Btree* createNode(char ch,Btree *l,Btree *r);//先序遍历void preOrder(Btree *BT){if(BT!= NULL){printf("%c",BT->data);preOrder(BT->left);preOrder(BT->right);}}//中序遍历void inOrder(Btree *BT){if(BT!= NULL){   inOrder(BT->left);printf("%c",BT->data);inOrder(BT->right);}}//后序遍历void postOrder(Btree *BT){if(BT!= NULL){   postOrder(BT->left);postOrder(BT->right);printf("%c",BT->data);}}//从括号表达式生成树void createTree(Btree **BT,char *str)//create a tree BT  according by hollow function str{     Btree *stack[maxSize],*p;     int top=-1,k,j=0;//top is the point of stack,k is the tab of son,j is the point of str;     char ch;     *BT=NULL;     ch=str[j];     while(ch!=NULL)     {          switch(ch)          {          case'(':top++;stack[top]=p;k=1;//left son,push to stack          break;          case')':top--;//pop stack          break;          case ',':k=2;//right son          break;          default:          /*p=(Btree *)malloc(sizeof(Btree));//set a node          if(!(p))          {               exit(0);          }          p->data=ch;          p->left=p->right=NULL;*/          p=createNode(ch,NULL,NULL);          if(*BT==NULL)//the root node          {               *BT=p;          }          else          {               switch(k)               {                 case 1:stack[top]->left=p;                 break;                 case 2:stack[top]->right=p;                 break;                 default:                 ;               }          }          }          j++;          ch=str[j];//get next char     }}/*void createTree(Btree **BT,char *str){     int j=0;//top is the point of stack,k is the tab of son,j is the point of str;     Btree *T;     char ch;     ch=str[j];     if(ch=='')  *BT=NULL;     else     {          if(!(T=))     }}*///生成节点Btree* createNode(char ch,Btree *l,Btree *r){     printf("Create now!\n");     Btree *p;     p=(Btree *)malloc(sizeof(Btree));//set a node     if(!(p))     {          exit(0);     }     p->data=ch;     p->left=l;     p->right=r;     return p;}//复制树//copy a tree and return the new tree's pointBtree* copyTree(Btree *T){     Btree *newptl,*newptr,*newT;     if(!T) return NULL;     if(T->left!=NULL)     {          newptl=copyTree(T->left);     }     else newptl=NULL;     if(T->right!=NULL) newptr=copyTree(T->right);     else newptr=NULL;     newT=createNode(T->data,newptl,newptr);     //newT=createNode(T->data,NULL,NULL);     //newT=NULL;     return newT;}//层次法打印树void disptree(Btree *BT){     Btree *stack[maxSize],*p;     int level[maxSize][2],top,n,i,width=4;     if(BT!=NULL)     {          printf("Display a tree by hollow means.\n");          top=1;          stack[top]=BT;//push root point to stack.          level[top][0]=width;          while(top>0)          {               p=stack[top];               n=level[top][0];               for(i=1;i<=n;i++)               printf(" ");               printf("%c",p->data);               for(i=n+1;i<maxWidth;i+=2)               printf("--");               printf("\n");               top--;               if(p->right!=NULL)               {                    top++;                    stack[top]=p->right;                    level[top][0]=n+width;                    level[top][1]=2;               }               if(p->left!=NULL)               {                    top++;                    stack[top]=p->left;                    level[top][0]=n+width;                    level[top][1]=1;               }          }     }}int TreeDepth(Btree *BT)//calculate the depth of tree{     int leftDep,rightDep;     if(BT==NULL) return 0;     else     {          leftDep=TreeDepth(BT->left);          rightDep=TreeDepth(BT->right);          if(leftDep>rightDep) return(leftDep+1);          else return (rightDep+1);     }}int NodeCount(Btree *BT)//cout the nodes{     if(BT==NULL) return 0;     else return (NodeCount(BT->left)+NodeCount(BT->right)+1);}int LeafCount(Btree *BT)//count the leafnodes{     if(BT==NULL) return 0;     else if(BT->left==NULL&&BT->right==NULL) return 1;     else return (LeafCount(BT->left)+LeafCount(BT->right));}void PrintTree(Btree *BT)//print tree{     if(BT!=NULL)     {          printf("%c",BT->data);          if(BT->left!=NULL||BT->right!=NULL)          {               printf("(");               PrintTree(BT->left);               if(BT->right!=NULL) printf(",");               PrintTree(BT->right);               printf(")");          }     }}main(){     Btree *B,*C;     char *s="A(B(D,E(H,I)),C(G))";     createTree(&B,s);     printf("preOder:");     preOrder(B);     printf("\n");     printf("inOder:");     inOrder(B);     printf("\n");     printf("postOder:");     postOrder(B);     printf("\n");     printf("Display tree\n");     disptree(B);     printf("copy tree\n");     //disptree(B);     C=copyTree(B);     preOrder(C);     disptree(C);     printf("Deep:  %d\n",TreeDepth(B));     printf("Sum of nodes:  %d\n",NodeCount(B));     printf("Sum of leafnodes:  %d\n",LeafCount(B));     PrintTree(B);printf("\nHello,world!");}