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
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| #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
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#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;
}