1162 Postfix Expression

分数 25

作者 陈越

单位 浙江大学

Given a syntax tree (binary), you are supposed to output the corresponding postfix 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 postfix expression, with parentheses reflecting the precedences of the operators.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)%))+)

  每个节点用一个结构体存储，结构体信息应包括结点值，左右孩子编号；
 * 处理步骤先找出根节点，再应根节点进行后序遍历，但应注意：如果有
 * 右孩子而没有左孩子，那么此根节点作为符号前缀，那么应该将根节点
 * 先于孩子结点输出，否则先左孩子再右孩子最后根节点；

/**
 * 每个节点用一个结构体存储，结构体信息应包括结点值，左右孩子编号；
 * 处理步骤先找出根节点，再应根节点进行后序遍历，但应注意：如果有
 * 右孩子而没有左孩子，那么此根节点作为符号前缀，那么应该将根节点
 * 先于孩子结点输出，否则先左孩子再右孩子最后根节点；
*/

#include <iostream>

using namespace std;

struct Node
{
    string s;
    int l, r;
};
const int N = 30;
bool hs[N]; //hs数组为了找出根节点
struct Node tr[N];
int n, root;

void Read()
{
    cin >> n;
    for(int i=1; i<=n; ++i)
    {
        string s;
        int l, r;
        cin >> s >> l >> r;
        
        if(l != -1) hs[l] = 1; 
        if(r != -1) hs[r] = 1;
        tr[i] = {s, l, r};
    }
}

void beh_traver(int root)
{
    if(root == -1)  return; //递归边界
    
    int l = tr[root].l, r = tr[root].r;
    string s = tr[root].s; 
    
    cout << '(';
    
    //如果有右孩子而没有左孩子，那么此根节点作为符号前缀，那么应该
    //将根节点先于孩子结点输出，否则先左孩子再右孩子最后根节点；
    if(l == -1 && r != -1)  cout << s; 
    
    beh_traver(l); 
    beh_traver(r);
    
    if(l == -1 && r != -1)  cout << ')';
    else cout << s << ')';
}

int main()
{
    Read();
    
    for(root = 1; hs[root] != 0; ++root); //找到根节点
    
    beh_traver(root);
    
    return 0;
}