Suffix expression (summer vacation daily question 4)

Given a binary expression tree , Please output the corresponding suffix expression , Brackets are required to reflect the priority of the operator .

Input format
The first line contains integers $N$, Represents the number of nodes . Node number $1\sim N$.

Next $N$ That's ok , Each row gives information about a node （ The first $i$ Row corresponds to the second row $i$ Nodes ）, The format is ：

data left_child right_child

among ,data It's not more than $10$ Character string ,left_child and right_child Are the numbers of the left and right child nodes of the node .

No child node （ namely NULL）, Then use −1 Express .

The following two figures correspond to the two examples given .

Output format
Output answers in one line , There must be no spaces between expression symbols .

Data range
$1\le N\le 20$
sample input 1：

8
* 8 7
a -1 -1
* 4 1
+ 2 5
b -1 -1
d -1 -1
- -1 6
c -1 -1

sample output 1：

(((a)(b)+)((c)(-(d))*)*)

sample input 2：

8
2.35 -1 -1
* 6 1
- -1 4
% 7 8
+ 2 3
a -1 -1
str -1 -1
871 -1 -1

sample output 2：

(((a)(2.35)*)(-((str)(871)%))+)

---

#include<iostream>
#include<cstring>

using namespace std;

const int N = 30;

int n;
string a[N];
int l[N], r[N];
bool st[N];

void dfs(int u){
    
    
    cout << '(';
    if(l[u] == -1 && r[u] != -1){
    
        cout << a[u];
        dfs(r[u]);
    }else{
    
        if(l[u] != -1) dfs(l[u]);
        if(r[u] != -1) dfs(r[u]);
        cout << a[u];
    }
    
    cout << ')';
}

int main(){
    
    
    cin >> n;
    
    for(int i = 1; i <= n; i++){
    
        cin >> a[i] >> l[i] >> r[i];
        if(l[i] != -1) st[l[i]] = true;
        if(r[i] != -1) st[r[i]] = true;
    }
    
    int head = -1;
    for(int i = 1; i <= n; i++)
        if(!st[i]) head = i;
    
    dfs(head);
    
    return 0;
}