数据结构实验之二叉树三：统计叶子数oj

数据结构实验之二叉树三：统计叶子数

Time Limit: 1000MS Memory Limit: 65536KB

Problem Description

已知二叉树的一个按先序遍历输入的字符序列，如abc,,de,g,,f,,, (其中,表示空结点)。请建立二叉树并求二叉树的叶子结点个数。

Input

连续输入多组数据，每组数据输入一个长度小于50个字符的字符串。

Output

输出二叉树的叶子结点个数。

Example Input

abc,,de,g,,f,,,

Example Output

3

Hint

 

Author

 xam

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct node
{
    char data;
    struct bitree *l,*r;
};
char a[100],b[55];//a前序b中序
int l1,h1,l2,h2;
int count;
struct node *creat() //前序建立
{
    struct node *root;
    char c;
    c = a[l1++];
    if(c == ',')
        return NULL;
    else
    {
        root = (struct node *)malloc(sizeof(struct node));
        root->data = c;
        root->l=creat();
        root->r=creat();
    }
    return root;
}
struct node *creat2(int n,char *a,char *b)//前序中序建立
{
    struct node *root;
    int i;
    if(n == 0)
        return NULL;
    root = (struct node *)malloc(sizeof(struct node));
    root->data=a[0];
    for(i = 0;i < n;i++)
    {
        if(b[i] == a[0])
            break;
    }
    root->l=creat2(i,a+1,b);
    root->r=creat2(n-i-1,a+i+1,b+i+1);

    return root;
}
struct node *creat3(int n,char *a,char *b)//中序后序建立
{
    struct node *root;
    int i;
    if(n == 0)
        return NULL;
    root = (struct node *)malloc(sizeof(struct node));
    root->data=b[n-1];
    for(i = 0;i < n;i++)
    {
        if(a[i] == b[n-1])
            break;
    }
    root->l=creat3(i,a,b);
    root->r=creat3(n-i-1,a+i+1,b+i);
    return root;
}
void preorder(struct node *root) //前序遍历
{
    struct node *p;
    p = root;
    if(p!=NULL)
    {
        printf("%c",p->data);
        preorder(p->l);
        preorder(p->r);
    }
}
void inorder(struct node *root)//中序遍历
{
    struct node *p;
    p = root;
    if(p!=NULL)
    {
        inorder(p->l);
        printf("%c",p->data);
        inorder(p->r);
    }
}
void postorder(struct node *root)//后序遍历
{
    struct node *p;
    p = root;
    if(p!=NULL)
    {
        postorder(p->l);
        postorder(p->r);
        printf("%c",p->data);
    }
}
void lorder(struct node *t)//层次遍历
{
    if(!t)          //考虑空树
        return ;
    struct node *q[100000],*p;
    int f,r;
    q[1] = t,f=r=1;
    while(f<=r)
    {
        p = q[f];
        f++;
        printf("%c",p->data);
        if(p->l!=NULL)
        {
            r++;
            q[r]=p->l;
        }
        if(p->r!=NULL)
        {
            r++;
            q[r]=p->r;
        }
    }
}
int treehight(struct node *t) //二叉数高度
{
    int h,lh,rh;
    if(t == NULL)
    {
        h = 0;
    }
    else
    {
        lh = treehight(t->l);
        rh = treehight(t->r);
        if(lh>rh)
            h = lh+1;
        else
            h = rh+1;
    }
    return h;
}
void leaf(struct node *t) //求叶子数
{
    if(t)
    {
        if((t->l==NULL)&&(t->r==NULL))
        {
            count++;
        }
        leaf(t->l);
        leaf(t->r);
    }
}
int main()
{
    int n,t;
    
    while(scanf("%s",a) != EOF){
    struct node *root;
    l1=0;
    count = 0;
    root = creat();
    leaf(root);
    printf("%d\n",count);}
    return 0;
}