7-10 树的遍历 (25分)

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;
}