A1130. 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))

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
typedef struct{
    int lchild, rchild;
    string data;
}node;
node tree[21];
int N, notRoot[21] = {0};
void inOrder(int root, int first){
    if(root == -1)
        return;
    string ss = tree[root].data;
    if(tree[root].rchild != -1 && first == 0)
        cout << "(";
    inOrder(tree[root].lchild, 0);
    cout << ss;
    inOrder(tree[root].rchild, 0);
    if(tree[root].rchild != -1 && first == 0)
        cout << ")";
}
int main(){
    scanf("%d", &N);
    for(int i = 1; i <= N; i++){
        int cl, cr;
        string ss;
        cin >> ss >> cl >> cr;
        tree[i].lchild = cl;
        tree[i].rchild = cr;
        tree[i].data = ss;
        if(cl != -1)
            notRoot[cl] = 1;
        if(cr != -1)
            notRoot[cr] = 1;
    }
    int root;
    for(root = 1; root <= N && notRoot[root] == 1; root++);
    inOrder(root, 1);
    cin >> N;
    return 0;
}

总结：

1、中序遍历即可，当该节点不是叶节点时，则需要先输出" ( "， 再递归访问左子树，右括号同理。起初我是用判断 + - * %来判断是否是非叶节点，结果最后一个测试点过不去。后来改成判断子树是否为空就全过了。这里可能是由于运算符号不止有+-*%,可能还有未知的运算符号。