1130 Infix Expression

1130 Infix Expression

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))

 总结：递归写了好久，就是没有写出来！！！多练吧

代码：

#include <iostream>
#include <vector>
using namespace std;

//根节点为运算符，叶节点为操作数
struct node{
    string s;
    int l,r;
};
vector<node> g;
bool has_root[50];
int root=1,n;
string dfs(int index){
    if(index==-1)   return "";
    if(g[index].r!=-1){//如果以当前点位根节点，右子树没有操作数的话，就说明这个当前点就是叶节点
        g[index].s = dfs(g[index].l) + g[index].s + dfs(g[index].r);
        if(index!=root) g[index].s = '(' + g[index].s + ')';
    }
    return g[index].s;
}
int main(){
    scanf("%d",&n);
    g.resize(n+1);
    
    for(int i=1;i<=n;i++){
        cin >> g[i].s >> g[i].l >> g[i].r;
        if(g[i].r!=-1)  has_root[g[i].r]=true;
        if(g[i].l!=-1)  has_root[g[i].l]=true;
    }
    while(has_root[root]) root++;
    cout << dfs(root) << endl;
    return 0;
}

好好学习，天天向上！

我要考研！