求二叉树的层次遍历oj

求二叉树的层次遍历

Time Limit: 1000MS Memory Limit: 65536KB

Problem Description

已知一颗二叉树的前序遍历和中序遍历，求二叉树的层次遍历。

Input

输入数据有多组，输入T，代表有T组测试数据。每组数据有两个长度小于50的字符串，第一个字符串为前序遍历，第二个为中序遍历。

Output

每组输出这颗二叉树的层次遍历。

Example Input

2
abc
bac
abdec
dbeac

Example Output

abc
abcde

Hint

 

Author

fmh

#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;
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;
}
int main()
{
    int n,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%s",a,b);
        struct node *root;
        n = strlen(a);
        root = creat2(n,a,b);
        lorder(root);
        printf("\n");
    }
    return 0;
}