用二叉連結串列儲存的二叉樹進行先序、中序和後序遍歷的演算法
用二叉樹連結串列作為儲存結構,完成二叉樹的建立,先序、中序和後序以及按層次遍歷的操作,求所有葉子及結點總數的操作
#include<iostream>
#include<cstdio>
#include<stdlib.h>
using namespace std;
typedef int Elemtype;
typedef struct BiTnode
{
Elemtype data;//資料域
struct BiTnode* Lchild,*Rchild; //左右子樹域;
}BiTnode,*BiTree;
int create(BiTree *T)
{
Elemtype ch;
Elemtype temp;
scanf("%d",&ch);
temp=getchar();
if(ch==-1)
{
*T=NULL;
}
else
{
*T=(BiTree)malloc(sizeof(BiTnode) );
if(!(*T))
{
exit(-1);
}
else
{
(*T)->data=ch;
printf("請輸入%d的左節點的值",ch);
create(&(*T)->Lchild);
printf("請輸入%d的右節點的值",ch);
create(&(*T)->Rchild);
}
}
return 1;
}
void Traverse(BiTree T)//前序遍歷二叉樹
{
if(NULL==T)
{
return;
}
else
{
printf("%d ",T->data);
Traverse(T->Lchild);
Traverse(T->Rchild);
}
}
//中序遍歷二叉樹
void midTraverse(BiTree T)
{
if(T==NULL){return;}
midTraverse(T->Lchild);
printf("%d ",T->data);
midTraverse(T->Rchild);
}
//後序遍歷二叉樹
void lasTraverse(BiTree T)
{
if(T==NULL){return;}
lasTraverse(T->Lchild);
lasTraverse(T->Rchild);
printf("%d ",T->data);
}
//求二叉樹的深度
int TreeDeep(BiTree T)
{
int deep=0;
if(T)
{
int leftdeep=TreeDeep(T->Lchild);
int rightdeep=TreeDeep(T->Rchild);
deep=leftdeep>=rightdeep?leftdeep+1:rightdeep+1;
}
return deep;
}
//求二叉樹葉子節點個數
int Leafcount(BiTree T,int &num)
{
if(T)
{
if(T->Lchild==NULL&&T->Rchild==NULL)
{
num++;
}
Leafcount(T->Lchild,num);
Leafcount(T->Rchild,num);
}
return num;
}
int main()
{
BiTree T;
BiTree *p=(BiTree*)malloc(sizeof(BiTree));
int deepth=0,num=0;
printf("請輸入第一個節點的值,-1代表沒有葉節點:\n");
create(&T);
printf("先序遍歷二叉樹:\n");
Traverse(T);
printf("\n");
printf("中序遍歷二叉樹:\n");
midTraverse(T);
printf("\n");
printf("後序遍歷二叉樹:\n");
lasTraverse(T);
printf("\n");
deepth=TreeDeep(T);
printf("樹的深度:%d\n",deepth);
printf("\n");
Leafcount(T,num);
printf("二叉樹的葉子節點個數為:%d\n",num);
printf("\n");
return 0;
}