# Solution by dilyar

• #### Approach #1 Tail-Recursive Solution [Accepted]

Algorithm

The key idea is to use `two pointer` and recursive method to help construct the BST.
We init the `root` node and then set its left and right subtree with corresponding indices.
Because of recursive calls on the `helper()` method, we will have a BST at the end.

Java

``````public class Solution {
/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public TreeNode sortedArrayToBST(int[] nums) {
if (nums == null || nums.length == 0) return null;
return helper(nums, 0, nums.length - 1);
}

private TreeNode helper(int[] nums, int left, int right) {
if (left > right) return null;
int mid = (left + right) / 2;
TreeNode node = new TreeNode(nums[mid]);
node.left = helper(nums, left, mid - 1);
node.right = helper(nums, mid + 1, right);
return node;
}
}
``````

Complexity Analysis

• Time complexity : \$\$O(n)\$\$, where n is the length of given array.

• Space complexity : \$\$O(n)\$\$. The recursive function is called `n` times in the method stack.

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