1130 Infix Expression～ (25分)

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))

难点：中序遍历加括号，加括号条件为，不是叶子节点且不是根节点输出括号，至于是"("还是")"，在之前输出"("，之后输出")"。

#include<iostream>

#include<vector>

using namespace std;

struct node{

    string data;
    int l, r;

};

struct tree{

    string data;
    tree * lchild;
    tree * rchild;

};
int node_cnt = 0;
tree * creat(tree * t,vector<node> &temp_vec, int start){
    
    if(start == -1){

        return NULL;
    }

    tree * root = new tree;
    node_cnt++;
    root -> data = temp_vec[start].data;

    root -> lchild = creat(root -> lchild, temp_vec, temp_vec[start].l);
    root -> rchild = creat(root -> rchild, temp_vec, temp_vec[start].r);

    return root;
}
string out_s = "";
tree * is_root;
void order_tree(tree * root){
    if(root == NULL)
        return;
    
    if((root -> lchild != NULL || root -> rchild != NULL) && is_root != root){
        printf("(");
        // out_s += "(";
    }
        
    order_tree(root -> lchild);
    // out_s += root -> data;
    printf("%s", root -> data.c_str());
    order_tree(root -> rchild);
    if((root -> lchild != NULL || root -> rchild != NULL) && is_root != root){
        printf(")");
        // out_s += ")";
    }
        
    
    
}
int main(){

    int n;
    cin >> n;
    vector<node> vec(n + 1);
    for(int i = 1; i <= n; i++){
        node temp;
        cin >> temp.data;
        scanf(" %d %d", &temp.l, &temp.r);
        vec[i] = temp;
    }

    // for(int i = 1; i <= n; i++){
    //     cout << vec[i].data << " " << vec[i].l << " " << vec[i].r << endl;
    // }
    tree * t;
    for(int i = 1; i <= n; i++){
        node_cnt = 0;
        t = NULL;
        t = creat(t ,vec, i);
        if(node_cnt == n){
            break;
        }

    }
    is_root = t;
    
    // cout << t -> data << " " << t -> lchild -> lchild -> data  << endl;
    order_tree(t);
    
    
    



    return 0;
}