PAT--1130. Infix Expression（中序加括号）

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.

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))

题解

中序加括号。

#include 
#include 
#include 
#include 
#include 
using namespace std;

const int maxn = 200 + 5;
struct Node{
    string val;
    int left, right;
};

Node nodes[maxn];
int n, root_num;
int numbers[maxn];

void infix(int root){
    if(root == -1) return;

    if(root != root_num && (nodes[root].left != -1 || nodes[root].right != -1)) cout << "(";
    infix(nodes[root].left);

    cout << nodes[root].val;

    infix(nodes[root].right);
    if(root != root_num && (nodes[root].left != -1 || nodes[root].right != -1)) cout << ")";

}

string infix1(int root){
    if(root == -1) return "";

    if(nodes[root].right != -1){
        nodes[root].val = infix1(nodes[root].left) + nodes[root].val + infix1(nodes[root].right);
        if(root != root_num) nodes[root].val = "(" + nodes[root].val + ")";
    }

    return nodes[root].val;
}

int main(){
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
#endif // ONLINE_JUDGE

    cin >> n;
    string val; int left, right;
    for(int i = 1; i <= n; ++i){
        cin >> val >> left >> right;
        nodes[i] = Node{val, left, right};
        if(left != -1)  numbers[left] = 1;
        if(right != -1) numbers[right] = 1;
    }

    // find the root of the tree
    for(int i = 1; i <= n; ++i) if(!numbers[i]) root_num = i;

    cout << infix1(root_num) << endl;
//    infix(root_num);
    return 0;
}