编写程序，实现以下功能： 1、对于给定的中缀表达式，创建与其对应的表达式树（二叉树），用二叉链表存储； 2、利用表达式树进行表达式求值。

好的，我可以给你提供一个C语言的实现，实现中缀表达式转换为表达式树并求值的功能。代码如下：

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

typedef struct TreeNode {
    char data;
    struct TreeNode *left;
    struct TreeNode *right;
} TreeNode;

typedef struct Stack {
    int top;
    int size;
    TreeNode **data;
} Stack;

Stack *createStack(int size) {
    Stack *S = (Stack *)malloc(sizeof(Stack));
    S->size = size;
    S->top = -1;
    S->data = (TreeNode **)malloc(sizeof(TreeNode *) * size);
    return S;
}

void push(Stack *S, TreeNode *x) {
    if (S->top == S->size - 1) {
        printf("Stack overflow\n");
        return;
    }
    S->data[++S->top] = x;
}

TreeNode *pop(Stack *S) {
    if (S->top == -1) {
        printf("Stack underflow\n");
        return NULL;
    }
    return S->data[S->top--];
}

TreeNode *createNode(char data) {
    TreeNode *newNode = (TreeNode *)malloc(sizeof(TreeNode));
    newNode->data = data;
    newNode->left = newNode->right = NULL;
    return newNode;
}

int isOperator(char data) {
    if (data == '+' || data == '-' || data == '*' || data == '/')
        return 1;
    return 0;
}

TreeNode *buildExpressionTree(char *expression) {
    Stack *S = createStack(strlen(expression));
    TreeNode *root = NULL;

    for (int i = 0; expression[i]; i++) {
        if (expression[i] == '(') {
            push(S, createNode(expression[i]));
        } else if (isdigit(expression[i])) {
            TreeNode *node = createNode(expression[i]);
            push(S, node);
        } else if (isOperator(expression[i])) {
            TreeNode *node = createNode(expression[i]);
            node->right = pop(S);
            node->left = pop(S);
            push(S, node);
        } else if (expression[i] == ')') {
            pop(S);
        }
    }

    if (!isEmpty(S)) {
        root = pop(S);
    }

    free(S->data);
    free(S);

    return root;
}

int evaluateExpressionTree(TreeNode *root) {
    if (!root) {
        return 0;
    }
    if (!root->left && !root->right) {
        return root->data - '0';
    }
    int leftValue = evaluateExpressionTree(root->left);
    int rightValue = evaluateExpressionTree(root->right);
    switch (root->data) {
        case '+': return leftValue + rightValue;
        case '-': return leftValue - rightValue;
        case '*': return leftValue * rightValue;
        case '/': return leftValue / rightValue;
    }
    return 0;
}

int main() {
    char expression[100];
    printf("Enter the expression: ");
    scanf("%s", expression);
    TreeNode *root = buildExpressionTree(expression);
    int result = evaluateExpressionTree(root);
    printf("Result: %d\n", result);
    return 0;
}

这个程序首先实现了一个栈和栈相关的操作函数，用于在转换中缀表达式为表达式树时辅助操作。然后，使用一个函数 `buildExpressionTree` 将中缀表达式转换为表达式树。最后，使用另一个函数 `evaluateExpressionTree` 对表达式树进行递归求值。