7-10 树的遍历 (25分)
#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<map>
#include<queue>
#include<iomanip>
#include<stack>
using namespace std;
#define STDIN freopen("in.in", "r", stdin);freopen("out.out", "w", stdout);
const int N = 10010;
int back[N];
int mid[N];
int le[N], ri[N];
int n;
int build(int l1, int r1, int l2, int r2)
{
if (l1 > r1) return 0;
int rt = back[r1];
int j;
for (int i = l2; i <= r2; i++){
if (mid[i] == rt){
j = i;
break;
}
}
int cnt = j - l2;
le[rt] = build(l1,l1+cnt-1, l2, j-1);
ri[rt] = build(l1+cnt, r1-1,j+1, r2);
return rt;
}
int main()
{
STDIN
cin >> n;
for (int i = 1; i <= n; i++) cin >> back[i];
for (int i = 1; i <= n; i++) cin >> mid[i];
le[0] = build(1,n,1,n);
queue<int> que;
vector<int> ans;
que.push(le[0]);
while (que.size())
{
auto t = que.front();
que.pop();
ans.push_back(t);
if (le[t] > 0) que.push(le[t]);
if (ri[t] > 0 ) que.push(ri[t]);
}
int len = ans.size();
for (int i = 0; i < len; i++)
{
printf("%d%c", ans[i], "
"[i == len-1]);
}
return 0;
}