PAT-A1130. 二叉树-中序遍历 变形输出

题目链接：https://www.patest.cn/contests/pat-a-practise/1130

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))

首先，数据输出的方式是中序遍历结果，然后考虑括号的位置。以图中的根结点为例，它的左右子树分别需要一对括号，而两个叶子结点之间是没有括号的，因为它们两个之间只有父结点的运算符，所以得出结论：如果一个结点，它的子结点还有子结点（也就是它的子结点不是叶子结点），那么这个子结点形成的子树需要用括号括起来。

#include <iostream>
#include <string>
#include <cstring>
using namespace std;

struct BTree{
    string data;
    int lchild, rchild;
};
BTree T[25];
bool bt[25];

void InOrder(int root)
{
    if(root == -1)
        return;
    else {
        if(T[root].lchild != -1 && (T[T[root].lchild].lchild != -1 || T[T[root].lchild].rchild != -1)){
            cout << "(";
            InOrder(T[root].lchild);
            cout << ")";
        }
        else
            InOrder(T[root].lchild);

        cout << T[root].data;

        if(T[root].rchild != -1 && (T[T[root].rchild].lchild != -1 || T[T[root].rchild].rchild != -1)){
            cout << "(";
            InOrder(T[root].rchild);
            cout << ")";
        }
        else
            InOrder(T[root].rchild);
    }
}

int main()
{
    int n,root;
    cin >> n;
    memset(bt, 0, sizeof(bt));
    for(int i = 1; i <= n; ++i){
        cin >> T[i].data >> T[i].lchild >> T[i].rchild;
        if(T[i].lchild != -1)
            bt[T[i].lchild] = true;
        if(T[i].rchild != -1)
            bt[T[i].rchild] = true;
    }
    for(int i = 1; i <= n; ++i){
        if(!bt[i]){
            root = i;
            break;
        }
    }

    InOrder(root);

    return 0;
}