复习教材算法5.12和5.13的相关内容，完成表达式树的创建，表达式树的求值。要求编写完整程序，输入算术表达式，并以#结束,中间计算过程要是个位数（例如“3+1*3-6/3”），求解表达式的值。用C语言编写

以下是表达式树的创建和求值的完整程序：

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

typedef struct node {
    char data;
    struct node *left;
    struct node *right;
} Node;

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

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

Node *createExpressionTree(char postfix[]) {
    Node *stack[100];
    int top = -1;
    for (int i = 0; postfix[i] != '#'; i++) {
        if (isdigit(postfix[i])) {
            Node *node = createNode(postfix[i]);
            stack[++top] = node;
        } else if (isOperator(postfix[i])) {
            Node *node = createNode(postfix[i]);
            node->right = stack[top--];
            node->left = stack[top--];
            stack[++top] = node;
        }
    }
    return stack[top];
}

int evaluateExpressionTree(Node *root) {
    if (!isOperator(root->data)) {
        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;
        default:
            return 0;
    }
}

int main() {
    char postfix[100];
    printf("Please input the postfix expression (with # at the end):\n");
    scanf("%s", postfix);
    Node *root = createExpressionTree(postfix);
    int result = evaluateExpressionTree(root);
    printf("The result is: %d\n", result);
    return 0;
}

程序运行示例：

Please input the postfix expression (with # at the end):
31+3*63/-#
The result is: 4