1.就算不知道用vector的初始化，也可以手动赋值创建子数组。

2.不断找到当前序列对应的根节点，计算他的子节点的总和，在这样递归处理过程中，注意要中序输出，所以对于是先遍历完左子树，然后输出答案，然后遍历右子树

#include <bits/stdc++.h>
using namespace std;
#define ll long long
//# define int long long
#define ull unsigned long long
#define pii pair<int,int>

#define baoliu(x, y) cout << fixed << setprecision(y) << x
#define endl  "\n"
#define debug1(x) cerr<<"x"<<" "
#define  debug2(x) cerr<<"x"<<endl
const int N =1100;
const int M = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
int n, m;
vector<int>ans;
void dfs(vector<int>pre,vector<int>in){
	if(pre.size()==0)return ;
	int root=pre[0];
	int pos=find(in.begin(),in.end(),root)-in.begin();
	//cerr<<pos<<endl;
	int s=0;
	vector<int>b1,b2;
	for(int i=0;i<pos;i++){
		b1.push_back(in[i]);
		s+=in[i];
	}
	for(int i=pos+1;i<in.size();i++){
		b2.push_back(in[i]);
		s+=in[i];
	}
	vector<int>pre1(pre.begin()+1,pre.begin()+b1.size()+1);
	vector<int>pre2(pre.begin()+b1.size()+1,pre.end());
	dfs(pre1,b1);
	ans.push_back(s);
	dfs(pre2,b2);
}
void solve(){
	cin>>n;
	vector<int>pre(n),in(n);
	for(int i=0;i<n;i++)cin>>pre[i];
	for(int i=0;i<n;i++)cin>>in[i];
	dfs(pre,in);
		//for(int i=0;i<n;i++)cerr<<ans[i]<<endl;
	for(int i=0;i<n;i++)cout<<ans[i]<<" ";
}
int main() {
    cin.tie(0);
    
    ios::sync_with_stdio(false);

    int t;
   //cin>>t;
     t=1;
    while (t--) {
solve();
    }
    return 0;
}
写法二：到叶子节点就返回，不到空结点，感觉不越界比较安全

#include <bits/stdc++.h>
using namespace std;
#define ll long long
//# define int long long
#define ull unsigned long long
#define pii pair<int,int>

#define baoliu(x, y) cout << fixed << setprecision(y) << x
#define endl “\n”
#define debug1(x) cerr<<”x”<<” “
#define debug2(x) cerr<<”x”<<endl
const int N =1100;
const int M = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
int n, m;
vectorans;
void dfs(vectorpre,vectorin){
if(pre.size()==1){ans.push_back(0);return;}
int root=pre[0];
int pos=find(in.begin(),in.end(),root)-in.begin();
//cerr<<pos<<endl;
int s=0;
vectorb1,b2;
for(int i=0;i<pos;i++){
b1.push_back(in[i]);
s+=in[i];
}
for(int i=pos+1;i<in.size();i++){
b2.push_back(in[i]);
s+=in[i];
}
vectorpre1(pre.begin()+1,pre.begin()+b1.size()+1);
vectorpre2(pre.begin()+b1.size()+1,pre.end());
dfs(pre1,b1);
ans.push_back(s);
dfs(pre2,b2);
}
void solve(){
cin>>n;
vectorpre(n),in(n);
for(int i=0;i<n;i++)cin>>pre[i];
for(int i=0;i<n;i++)cin>>in[i];
dfs(pre,in);
//for(int i=0;i<n;i++)cerr<<ans[i]<<endl;
for(int i=0;i<n;i++)cout<<ans[i]<<” “;
}
int main() {
cin.tie(0);

ios::sync_with_stdio(false);

int t;

//cin>>t;
t=1;
while (t–) {
solve();
}
return 0;
}