1162 Postfix Expression (25 point(s)) PAT

1162 Postfix Expression (25 point(s))

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题目

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.

infix1.JPG infix2.JPG
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)%))+)

思路

题意

给出一个语法树，按照后缀表达输出。输入n个结点，按照结点值，左孩子index，右孩子index的格式读入（结点读入的次序即其index）。输出格式为每个表达式都用（）括起来表示优先级。遇到结点无左孩子有右孩子的情况，表达式变为前缀表达。

思路

用tree[][2]建立index树，用node[]记录每个index对应的结点值。读入输入的时候建树，然后遍历树的时候输出。遍历树分三种情况：

1. 结点同时有左右孩子，后序遍历输出
2. 结点无左孩子有右孩子，先输出结点值再遍历右子树
3. 结点无孩子，直接输出结点值。

注意每次遍历的头尾要输出(和)

解法

#include<bits/stdc++.h>
using namespace std;

int n, root;
int tree[25][2], visited[25] = { 0 };
string node[25];

void traverse(int index) {//左1右1，左0右1，左0右0
	printf("(");
	if (tree[index][0] > 0) {//左1
		traverse(tree[index][0]);//左子树
		traverse(tree[index][1]);//右子树
		printf("%s", node[index].c_str());
	}
	else if (tree[index][1] > 0) {//左0右1
		printf("%s", node[index].c_str());//输出结点本身
		traverse(tree[index][1]);//右子树
	}
	else {//左0右0
		printf("%s", node[index].c_str());//输出结点本身
	}
	printf(")");
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		string data;
		cin >> data;
		node[i] = data;
		int left, right;
		scanf("%d%d", &left, &right);
		tree[i][0] = left;
		tree[i][1] = right;
		visited[left] = visited[right] = 1;
	}
	for (int i = 1; i <= n; i++)
		if (visited[i] == 0) {
			root = i;
			break;
		}
	traverse(root);
	printf("\n");
	return 0;
}

注意

• 当左孩子无右孩子有的时候均是特殊表达