A question about how to judge an array is the back sequence of a binary tree

#include <stdio.h>
#include <stdlib.h>
typedef struct btree bt;
//二叉树的节点结构定义
struct btree {
    int _val;
    int _count;//此处优化结构,当元素重复时将该节点的计数加一
    bt *_lc;
    bt *_rc;
};
/**
 * 挂载节点
 * @param head 二叉树的根节点的指针
 * @param val 待插叙节点的值
 */
void insert(bt *const head, int const val) {
    bt *node = malloc(sizeof(bt));
    node->_val = val;
    node->_count = 1;
    node->_lc = NULL;
    node->_rc = NULL;
    bt *tmp = head;
    while (1) {
        if (tmp->_val == val) {
            tmp->_count++;
            return;
        } else if (tmp->_val > val) {
            if (tmp->_lc == NULL)
                break;
            else
                tmp = tmp->_lc;
        } else {
            if (tmp->_rc == NULL)
                break;
            else
                tmp = tmp->_rc;
        }
    }
    if (tmp->_val > val)
        tmp->_lc = node;
    else
        tmp->_rc = node;
}
/**
 * 构建二叉树
 * @param arr 构建二叉树的输入-数组
 * @param len 数组的长度
 * @return 二叉树根节点的指针
 */
bt *buildTree(int const * const arr, size_t const len) {
    bt *head = malloc(sizeof(bt));
    head->_val = *(arr + len - 1);
    head->_count = 1;
    head->_lc = NULL;
    head->_rc = NULL;
    int i;
    for (i = len - 2; i >= 0; --i) {
        insert(head, *(arr + i));
    }
    return head;
}
/**
 * 中序遍历(采用的递归算法)
 * @param node 当前的"根"节点指针
 */
void travelMidSeq(bt const * const node) {
    if (node->_lc != NULL)
        travelMidSeq(node->_lc);
    int i;
    int len = node->_count;
    for (i = 0; i < len; ++i)
        printf("%d ", node->_val);
    if (node->_rc != NULL)
        travelMidSeq(node->_rc);
}
//记录当前已遍历元素的个数;
int index = 0;
//标记当前是否找到错误元素并记录错误元素的位置
char flag = -1;
/**
 * 释放资源(递归算法)---实质就是后续遍历的算法,并校验数组中元素是不是后续遍历的顺序
 * @param node 当前需要处理的节点指针
 * @param arr 要排序的数组的指针
 */
void releaseTreeAndCheck(bt *node, int const * const arr) {
    if (node->_lc != NULL)
        releaseTreeAndCheck(node->_lc, arr);
    if (node->_rc != NULL)
        releaseTreeAndCheck(node->_rc, arr);
    if (flag < 0) {
        if (*(arr + index) != node->_val)
            flag = index;
        else
            index++;
    }
    printf("%d ", node->_val);
    free(node);
}
/**
 * 打印数组
 * @param arr 数组的指针
 * @param len 数组的长度
 */
void print(int const * const arr, size_t const len) {
    int i;
    printf("original sequence:\n");
    for (i = 0; i < len; ++i)
        printf("%d ", *(arr + i));
}
int main(void) {
    int arr[] = {1, 3, 2, 5, 7, 9, 4, 9, 11, 10, 13, 15, 14, 12, 8};
    //int arr[] = {1, 3, 2, 5, 7, 6, 4, 9, 11, 10, 13, 15, 14, 12, 8};
    size_t len = sizeof(arr) / sizeof(int);
    print(arr, len);
    bt *tree = buildTree(arr, len);
    printf("\nmiddle sequence travel result :\n");
    travelMidSeq(tree);
    printf("\nback sequence travel result :\n");
    releaseTreeAndCheck(tree, arr);
    if (flag >= 0) printf("\nsorry! your input seems not the back sequence of a binary tree...\nthe wrong member from index %d", flag);
    else printf("\ncongratulations! your input is the back sequence of a binary tree...");
    return 0;
}