SDUTOJ 1489 求二叉树的先序遍历

标签：二叉树

<img src="http://img.blog.csdn.net/20141014212549703?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvUl9NaXNheWE=/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center" alt="" />

#include<stdio.h>
#include<string.h>
int cont;
char st1[100],st2[100],ans[100];
void build(int n,char *s1,char *s2,char *s)
{
    if(n<=0) return;
    int p=strchr(s1,s2[n-1])-s1;
	ans[cont++]=s2[n-1];
    build(p,s1,s2,s);
    build(n-p-1,s1+p+1,s2+p,s+p);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        cont=0;
        scanf("%s%s",st1,st2);
        int len=strlen(st1);
        build(len,st1,st2,ans);
        ans[len]='\0';
        printf("%s\n",ans);
    }
	return 0;
}

SDUTOJ 1489 求二叉树的先序遍历

标签：二叉树

原文地址：http://blog.csdn.net/r_misaya/article/details/40085127