Convert Sorted Array to Binary Search Tree & Convert Sorted List to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

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要点就是找到中心点，然后分别递归构造左边的数和右边的数

TreeNode* sortedArrayToBST(vector<int>& nums, int beg, int end) {
    if (beg > end)
        return nullptr;
    int mid = (beg + end) >> ;
    TreeNode* root = new TreeNode(nums[mid]);
    root->left = sortedArrayToBST(nums, beg, mid - );
    root->right = sortedArrayToBST(nums, mid + , end);
    return root;

}

TreeNode* sortedArrayToBST(vector<int>& nums) {
    return sortedArrayToBST(nums, , nums.size()-);
}

链表为了代码的简洁，我使用了二级指针，不过可读性变差了

TreeNode* sortedListToBST(ListNode* head) {
    if (head == nullptr)
        return nullptr;
    ListNode* fast = head;
    ListNode** slow = &head;
    while (fast->next != nullptr && fast->next->next != nullptr)
    {
        fast = fast->next->next;
        slow = &((*slow)->next);
    }
    TreeNode* root = new TreeNode((*slow)->val);

    root->right = sortedListToBST((*slow)->next);
    *slow = nullptr;
    root->left = sortedListToBST(head);

    return root;
}