试设计一个判别给定二叉树是否为二叉排序树的算法。用c语言

以下是用C语言实现判断给定二叉树是否为二叉排序树的算法：

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

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

int preVal = -2147483648; // 二叉树中序遍历的前一个节点的值，初始化为int类型的最小值

int isValidBST(TreeNode* root) {
    if (root == NULL) { // 空树是二叉排序树
        return 1;
    }
    if (!isValidBST(root->left)) { // 遍历左子树
        return 0;
    }
    if (root->val <= preVal) { // 如果当前节点的值小于等于前一个节点的值，说明不是二叉排序树
        return 0;
    }
    preVal = root->val; // 更新前一个节点的值
    if (!isValidBST(root->right)) { // 遍历右子树
        return 0;
    }
    return 1; // 是二叉排序树
}

int main() {
    // 构造一棵二叉排序树
    TreeNode *root = (TreeNode*)malloc(sizeof(TreeNode));
    root->val = 4;
    root->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->val = 2;
    root->left->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->left->val = 1;
    root->left->left->left = NULL;
    root->left->left->right = NULL;
    root->left->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->right->val = 3;
    root->left->right->left = NULL;
    root->left->right->right = NULL;
    root->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->val = 6;
    root->right->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->left->val = 5;
    root->right->left->left = NULL;
    root->right->left->right = NULL;
    root->right->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->right->val = 7;
    root->right->right->left = NULL;
    root->right->right->right = NULL;

    if (isValidBST(root)) {
        printf("给定二叉树是二叉排序树\n");
    } else {
        printf("给定二叉树不是二叉排序树\n");
    }

    return 0;
}