剑指OFFER之重建二叉树（九度OJ1385）

8
1 2 4 7 3 5 6 8
4 7 2 1 5 3 8 6
8
1 2 4 7 3 5 6 8
4 1 2 7 5 3 8 6

7 4 2 5 8 6 3 1 
No

#include <stdio.h>
#include <stdlib.h>
#include <memory.h>
int findRoot(int *arr1,int begin1,int end1,int *arr2,int begin2,int end2,int *final);
int final[1000];
int flag = 999;
int main(void){
    int n,i;
    int arr1[1000];
    int arr2[1000];
    while(scanf("%d",&n) != EOF && n <= 1000 && n >= 1){
        //initialize
        memset(arr1,0,sizeof(int)*1000);
        memset(arr2,0,sizeof(int)*1000);
        memset(final,0,sizeof(int)*1000);
        flag = 999;
        //input
        for(i=0;i<n;i++)
            scanf("%d",&arr1[i]);
        for(i=0;i<n;i++)
            scanf("%d",&arr2[i]);

        if(findRoot(arr1,0,n-1,arr2,0,n-1,final) == 0)
            printf("No\n");
        else{
            for(i=flag+1;i<1000;i++)
                printf("%d ",final[i]);
            printf("\n");
        }
    }
    return 0;
}
int findRoot(int *arr1,int begin1,int end1,int *arr2,int begin2,int end2,int *final){
    int i,j;
    int sum = 0;
    if(begin1==end1 && begin2 == end2){
        final[flag] = arr1[begin1];
        flag--;
        return 1;
    }
    for(i=begin1 ; i<=end1;i++){
        for(j=begin2 ; j <=end2 ; j++){
            if(arr1[i] == arr2[j])
                sum++;
        }
    }
    if(sum != (end1-begin1+1) && sum != (end2 - begin2+1)){
        return 0;
    }

    final[flag] = arr1[begin1];
    flag--;

    int numberofRoot = -1;
    for(i=begin2 ; i<=end2 ; i++){
        if(arr1[begin1] == arr2[i]){
            numberofRoot = i;
            //printf("找到跟在arr2的坐标为%d\n",numberofRoot);
            break;
        }
    }
    if(numberofRoot != end2){
        //printf("right %d %d %d %d\n",begin1+numberofRoot-begin2+1,end1,numberofRoot+1,end2);
        if(!findRoot(arr1,begin1+numberofRoot-begin2+1,end1,arr2,numberofRoot+1,end2,final)){
            return 0;
        }
    }
    if(numberofRoot != begin2){
        //printf("left %d %d %d %d\n",begin1+1,begin1+numberofRoot-begin2,begin2,numberofRoot-1);
        if(!findRoot(arr1,begin1+1,begin1+numberofRoot-begin2,arr2,begin2,numberofRoot-1,final)){//左子树
            return 0;
        }
    }
    return 1;
}