写一个C语言代码，利用表达式树将中缀表达式转为后缀表达式

好的，请等待我一会，我会为您生成这段代码。

以下是用C语言编写的中缀表达式转后缀表达式的代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_STACK_SIZE 100

typedef struct {
    char data[MAX_STACK_SIZE];
    int top;
} Stack;

void init_stack(Stack *s) {
    s->top = -1;
}

void push(Stack *s, char c) {
    if (s->top == MAX_STACK_SIZE - 1) {
        printf("Stack overflow!\n");
        exit(0);
    }

    s->top++;
    s->data[s->top] = c;
}

char pop(Stack *s) {
    if (s->top == -1) {
        printf("Stack underflow!\n");
        exit(0);
    }

    char c = s->data[s->top];
    s->top--;

    return c;
}

char peek(Stack *s) {
    if (s->top == -1) {
        printf("Stack underflow!\n");
        exit(0);
    }

    return s->data[s->top];
}

int is_operator(char c) {
    return c == '+' || c == '-' || c == '*' || c == '/';
}

int precedence(char c) {
    switch (c) {
        case '+':
        case '-':
            return 1;
        case '*':
        case '/':
            return 2;
        default:
            return 0;
    }
}

char *infix_to_postfix(char *infix) {
    Stack s;
    char *postfix = (char *)malloc(sizeof(char) * (strlen(infix) + 1));
    int i, j;

    init_stack(&s);
    j = 0;

    for (i = 0; infix[i] != '\0'; i++) {
        if (!is_operator(infix[i])) {
            postfix[j] = infix[i];
            j++;
        } else {
            while (s.top >= 0 && precedence(peek(&s)) >= precedence(infix[i])) {
                postfix[j] = pop(&s);
                j++;
            }
            push(&s, infix[i]);
        }
    }

    while (s.top >= 0) {
        postfix[j] = pop(&s);
        j++;
    }

    postfix[j] = '\0';

    return postfix;
}

int main() {
    char infix[100], *postfix;

    printf("Enter an infix expression: ");
    scanf("%s", infix);

    postfix = infix_to_postfix(infix);

    printf("Postfix expression: %s\n", postfix);

    return 0;
}

您可以通过输入一个中缀表达式，得到对应的后缀表达式。