PAT - A급 - 1130.Infix Expression(25)

제목 설명:
Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the precedences of the operators.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N ( <= 20 ) which is the total number of nodes in the syntax tree. Then N lines follow, each gives the information of a node (the i-th line corresponds to the i-th node) in the format:
data left_child right_child
where data is a string of no more than 10 characters, left_child and right_child are the indices of this node's left and right children, respectively. The nodes are indexed from 1 to N. The NULL link is represented by -1. The figures 1 and 2 correspond to the samples 1 and 2, respectively.
Figure 1
Figure 2
Output Specification:
For each case, print in a line the infix expression, with parentheses reflecting the precedences of the operators. Note that there must be no extra parentheses for the final expression, as is shown by the samples. There must be no space between any symbols.
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))

제목 사고방식:
접두사 표현식 두 갈래 트리를 지정하고 접두사 표현식을 출력합니다.주의해야 할 것은 괄호(parentheses)의 출력입니다. 두 갈래 나무의 뿌리 노드와 잎 결점은 괄호를 출력할 필요가 없습니다.절차에 제한을 가하면 된다.
제목 코드:

#include
#include
#include
#include
#define MAXN 21
using namespace std;
int n, Root;
int vis[MAXN];

struct Node{
	string s;
	int l, r;
}node[MAXN];


void dfs(int root)
{	//border 
	if(root==-1) return ;
	//  
	if(root!=Root&&(node[root].l!=-1||node[root].r!=-1)) cout<>node[i].s>>node[i].l>>node[i].r;
		vis[node[i].l] = vis[node[i].r] = 1;
	}
	//  
	Root = 1;
	while(vis[Root])Root++;
	
	dfs(Root);
	
	return 0;
}