# LC 108. Convert Sorted Array to Binary Search Tree

### [Problem link](https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/)

###### tags: `leedcode` `python` `c++` `easy` `BST`

Given an integer array <code>nums</code> where the elements are sorted in **ascending order** , convert it to a **height-balanced** binary search tree.

**Example 1:**
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/btree1.jpg" style="width: 302px; height: 222px;" />

Input: nums = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: [0,-10,5,null,-3,null,9] is also accepted:

<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/btree2.jpg" style="width: 302px; height: 222px;" />

**Example 2:**
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/btree.jpg" style="width: 342px; height: 142px;" />

Input: nums = [1,3]
Output: [3,1]
Explanation: [1,null,3] and [3,1] are both height-balanced BSTs.

**Constraints:**

- <code>1 <= nums.length <= 10⁴</code>
- <code>-10⁴ <= nums[i] <= 10⁴</code>
- <code>nums</code> is sorted in a **strictly increasing** order.

## Solution 1 - BST
#### Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
        if not nums:
            return None

        mid = len(nums) // 2
        root = TreeNode(val=nums[mid])

        root.left = self.sortedArrayToBST(nums[:mid])
        root.right = self.sortedArrayToBST(nums[mid + 1:])

        return root

#### C++

class Solution {
public:
    TreeNode *traversal(vector<int> &nums, int left, int right) {
        if (left > right) {
            return nullptr;
        }
        int mid = left + ((right - left) / 2);
        TreeNode *cur = new TreeNode(nums[mid]);
        cur->left = traversal(nums, left, mid - 1);
        cur->right = traversal(nums, mid + 1, right);
        return cur;
    }

    TreeNode* sortedArrayToBST(vector<int>& nums) {
        TreeNode *root = traversal(nums, 0, nums.size() - 1);
        return root;
    }
};

>### Complexity
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(1) |

## Note
follow up: [LC 1382. Balance a Binary Search Tree](https://hackmd.io/@Alone0506/Sk6V3_k3n)