1020. Tree Traversals (25)_2018_3_24

1020. Tree Traversals (25)

时间限制 400 ms
内存限制 65536 kB
代码长度限制 16000 B
判题程序 Standard 作者 CHEN, Yue
Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:

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

Sample Output:

4 1 6 3 5 7 2

提交代码#include<bits/stdc++.h>
using namespace std;

const int N(35);
int a
,b
;
int cnt;

struct AA{
int a;
int val;
AA(){}
AA(int aa,int vv):a(aa),val(vv){}
};

AA ans
;
int cmp(AA a,AA b){
return a.val<b.val;
}

void solve(int la,int ra,int lb,int rb,int w){
if(la>ra)return;
if(la==ra){
ans[cnt++]=AA(a[ra],w);
return;
}
ans[cnt++]=AA(a[ra],w);
int i=lb;
for(;i<=rb;i++)
if(a[ra]==b[i])break;
solve(la,la+i-lb-1,lb,i-1,w+1);
solve(la+i-lb,ra-1,i+1,rb,w+1);
}

int main(){
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int i=0;i<n;i++)
scanf("%d",&b[i]);
cnt=0;
solve(0,n-1,0,n,0);
stable_sort(ans,ans+cnt,cmp);
for(int i=0;i<cnt;i++)
if(i==0)printf("%d",ans[i].a);
else printf(" %d",ans[i].a);
}