#include <stdio.h> #include <stdlib.h> #include <ctype.h> #include <string.h> typedef struct TreeNode { char data; struct TreeNode* left; struct TreeNode* right; } TreeNode; TreeNode* createNode(char data) { TreeNode* node = (TreeNode*) malloc(sizeof(TreeNode)); node->data = data; node->left = NULL; node->right = NULL; return node; } TreeNode* createBinaryTree(char* expression, int start, int end) { if (start > end) { return NULL; } TreeNode* root = NULL; int i; int flag = 0; for (i = start; i <= end; i++) { if (expression[i] == '(') { flag++; } else if (expression[i] == ')') { flag--; } else if (flag == 0 && (expression[i] == '+' || expression[i] == '-' || expression[i] == '*' || expression[i] == '/')) { root = createNode(expression[i]); break; } } if (root == NULL) { for (i = start; i <= end; i++) { if (isdigit(expression[i])) { root = createNode(expression[i]); break; } } } root->left = createBinaryTree(expression, start, i - 1); root->right = createBinaryTree(expression, i + 1, end); return root; } int evaluate(TreeNode* root) { if (root == NULL) { return 0; } if (root->left == NULL && root->right == NULL) { return root->data - '0'; } int leftValue = evaluate(root->left); int rightValue = evaluate(root->right); switch (root->data) { case '+': return leftValue + rightValue; case '-': return leftValue - rightValue; case '*': return leftValue * rightValue; case '/': return leftValue / rightValue; default: return 0; } } int main() { char expression[100]; printf("请输入中序表达式:"); scanf("%s", expression); TreeNode* root = createBinaryTree(expression, 0, strlen(expression) - 1); int result = evaluate(root); printf("表达式的值为:%d\n", result); return 0; }给每一行加上详细注释,并说明使用了什么方法编写的代码,以及用这种方法的好处
时间: 2024-01-27 21:04:49 浏览: 28
这段代码使用了二叉树和递归的方法,实现了对中序表达式的求值。
注释如下:
```c
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
// 定义二叉树结点
typedef struct TreeNode {
char data; // 结点数据
struct TreeNode* left; // 左子树
struct TreeNode* right; // 右子树
} TreeNode;
// 创建二叉树结点
TreeNode* createNode(char data) {
TreeNode* node = (TreeNode*) malloc(sizeof(TreeNode));
node->data = data;
node->left = NULL;
node->right = NULL;
return node;
}
// 创建二叉树
TreeNode* createBinaryTree(char* expression, int start, int end) {
if (start > end) {
return NULL;
}
TreeNode* root = NULL;
int i;
int flag = 0; // 记录左右括号的数量差
// 寻找运算符
for (i = start; i <= end; i++) {
if (expression[i] == '(') {
flag++;
} else if (expression[i] == ')') {
flag--;
} else if (flag == 0 && (expression[i] == '+' || expression[i] == '-' || expression[i] == '*' || expression[i] == '/')) {
root = createNode(expression[i]);
break;
}
}
// 如果没有找到运算符,说明是数字,创建相应结点
if (root == NULL) {
for (i = start; i <= end; i++) {
if (isdigit(expression[i])) {
root = createNode(expression[i]);
break;
}
}
}
// 递归创建左右子树
root->left = createBinaryTree(expression, start, i - 1);
root->right = createBinaryTree(expression, i + 1, end);
return root;
}
// 计算二叉树表达式的值
int evaluate(TreeNode* root) {
if (root == NULL) {
return 0;
}
if (root->left == NULL && root->right == NULL) {
return root->data - '0';
}
int leftValue = evaluate(root->left);
int rightValue = evaluate(root->right);
// 根据运算符计算表达式的值
switch (root->data) {
case '+':
return leftValue + rightValue;
case '-':
return leftValue - rightValue;
case '*':
return leftValue * rightValue;
case '/':
return leftValue / rightValue;
default:
return 0;
}
}
int main() {
char expression[100];
printf("请输入中序表达式:");
scanf("%s", expression);
TreeNode* root = createBinaryTree(expression, 0, strlen(expression) - 1);
int result = evaluate(root);
printf("表达式的值为:%d\n", result);
return 0;
}
```
使用二叉树和递归的方法,可以将中序表达式转化为二叉树,并通过递归遍历二叉树求出表达式的值。这种方法的好处是简单易懂,代码实现简单,易于调试和修改。同时,对于复杂的表达式,使用二叉树和递归的方法也能够高效地求出表达式的值。