将有序数组转换为二叉搜索树

[dql870527]

报错信息：
Line 22: Char 21: runtime error: load of misaligned address 0xbebebebebebebebe for type 'int', which requires 4 byte alignment (solution.c)
0xbebebebebebebebe: note: pointer points here
<memory cannot be printed>

[寻开心]

贴代码， 谁知道你写的是啥啊

[Logy_S]

引用 1 楼 寻开心 的回复:

贴代码， 谁知道你写的是啥啊

哈哈，他可能让你翻译翻译。哈哈哈

[RichAndMerry]

struct TreeNode* fuzhi(int* nums, int left, int right);
struct TreeNode* sortedArrayToBST(int* nums, int numsSize){

    if(numsSize == 0)
        return NULL;
    return fuzhi(nums, 0, numsSize - 1);
}
struct TreeNode* fuzhi(int* nums, int left, int right)
{
    if(left > right || left < 0)
        return NULL;
    struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    int mid = left + (right - left) / 2;
    root->val = nums[mid];
    root->left = fuzhi(nums, left, mid - 1);
    root->right = fuzhi(nums, mid + 1, right);
    return root;
}