1130. Infix Expression (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 1Figure 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 7a -1 -1* 4 1+ 2 5b -1 -1d -1 -1- -1 6c -1 -1

Sample Output 1:

(a+b)*(c*(-d))

Sample Input 2:

82.35 -1 -1* 6 1- -1 4% 7 8+ 2 3a -1 -1str -1 -1871 -1 -1

Sample Output 2:

(a*2.35)+(-(str%871))

题意：给出二叉树，按中序遍历输出（带括号）

思路：因为输入的结构就是二叉树了，所以不需要自己再写一个（只要模拟就好了）

           关键点在于，要找出根节点 （没有被左右子树提到的就是根节点）

代码：

#include<string>#include<iostream>using namespace std;struct node {string data;int lchild;int rchild;}num[30];void infix(int root){if (root == -1) return;if (num[root].lchild > 0 || num[root].rchild > 0)cout << "(";if (num[root].lchild > 0) infix(num[root].lchild);cout << num[root].data;if (num[root].rchild > 0) infix(num[root].rchild);if (num[root].lchild > 0 || num[root].rchild > 0)cout << ")";}int main(){int n,root;cin >> n;int check[30] = { 0 };for (int i = 1; i <= n; i++){cin >> num[i].data >> num[i].lchild >> num[i].rchild;if (num[i].lchild > 0) check[num[i].lchild] = 1;if (num[i].rchild > 0) check[num[i].rchild] = 1;}for (root = 1; root <= n; root++)if (!check[root]) break;infix(num[root].lchild);cout << num[root].data;infix(num[root].rchild);cout << endl;}

结尾： 这题做的比较顺利，但是交上去的时候后面几个测试点一直提示段错误

    仔细想了想，因为我输出是 左子树 右子树 分开的，而前面没有加判断>0 的 if，所以最后加了第11行 如果==-1 就 retur