1162 Postfix Expression（29行+超详细注释）

分数 25

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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)%))+)

代码长度限制

16 KB

时间限制

400 ms

内存限制

64 MB

 注意：其中+，-两个符号既可以表示加减也可以表示正负，当表示正负的时候该节点一定是没有左孩子的

#include<bits/stdc++.h>
using namespace std;
const int N=25;
int n,l[N],r[N],exist[N];
string w[N];
string post(int u){
    string left,right;
    if(l[u]==-1&&r[u]==-1)return w[u];//若是根结点则返回该结点符号 
    if(l[u]!=-1&&r[u]!=-1){//若左右孩子都有 
        left="("+post(l[u])+")";//记录左子树及加上对应括号 
        right="("+post(r[u])+")";//记录右子树及加上对应的括号 
        return left+right+w[u];//返回左孩子，右孩子和该节点的符号 
    }
    else if(w[u]=="-"||w[u]=="+"&&l[u]==-1){//若该节点是+，-两个符号且没有左孩子，说明两个符号表示正负 
        right="("+post(r[u])+")";//只用遍历右子树并记录即可 
        return w[u]+right;//返回该符号和右子树 
    }
}
int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>w[i]>>l[i]>>r[i];
        exist[l[i]]=exist[r[i]]=1;//记录出现过的结点 
    }
    int root;
    for(int i=1;i<=n;i++)if(!exist[i])root=i;//没出现的结点是根结点 
    cout<<"("+post(root)+")";
    return 0;
}