假设二叉树每个结点的元素均为一个单字符，根据给定的字符序列按照先序遍历的顺序递归创建该树的二叉链表，然后判断该二叉树是否为二叉排序树。,用C语言写,写的简单点

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

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

// 先序遍历创建二叉树
TreeNode* createTree(char* str, int* index) {
    if (str[*index] == '#') {
        (*index)++;
        return NULL;
    }
    TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
    node->val = str[*index];
    (*index)++;
    node->left = createTree(str, index);
    node->right = createTree(str, index);
    return node;
}

// 判断是否为二叉排序树
bool isBST(TreeNode* root, char min, char max) {
    if (root == NULL) {
        return true;
    }
    if (root->val <= min || root->val >= max) {
        return false;
    }
    return isBST(root->left, min, root->val) && isBST(root->right, root->val, max);
}

int main() {
    char str[] = "FBD##E##A##C##";
    int index = 0;
    TreeNode* root = createTree(str, &index);
    if (isBST(root, 'A'-1, 'Z'+1)) {
        printf("该二叉树是二叉排序树\n");
    } else {
        printf("该二叉树不是二叉排序树\n");
    }
    return 0;
}