PAT A1130. Infix Expression (25)

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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))

解答思路：

1.用静态方法建立二叉树，并确定二叉树头结点（建立一个bool数组进行判断，若标号i作为左右子树出现过，则不为头结点）

2.中序遍历二叉树。

2.1若结点不为叶子结点（右子树不为空），且不是头结点（题目要求最外侧无括号），输出左括号

2.2遍历左子树；输出data；遍历右子树；

2.3若结点不为叶子结点（右子树不为空），且不是头结点（题目要求最外侧无括号），输出右括号

代码：

#include <stdio.h>#define MAX 21using namespace std;struct Node{char data[11];int left, right;}node[MAX];int n, root;bool judge[MAX] = {false};void inorder(int root, int level){if(root != -1){if(node[root].right != -1 && level > 0) printf("(");inorder(node[root].left, level+1);printf("%s", node[root].data);inorder(node[root].right, level+1);if(node[root].right != -1 && level > 0)printf(")");}}int main(){scanf("%d", &n);for(int i = 1; i <= n; i++){scanf("%s %d %d", node[i].data, &node[i].left, &node[i].right);if(node[i].left != -1) judge[node[i].left] = true;if(node[i].right != -1) judge[node[i].right] = true;}for(int i = 1; i <= n; i++){if(!judge[i]){root = i;break;}}inorder(root, 0);return 0;}

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