108. Convert Sorted Array to Binary Search Tree

https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {number[]} nums
 * @return {TreeNode}
 */
function sortedArrayToBST(nums) {
  if (!nums.length) {
    return null
  }

  const mid = nums.length / 2 >> 0
  const root = new TreeNode(nums[mid])
  root.left = sortedArrayToBST(nums.slice(0, mid))
  root.right = sortedArrayToBST(nums.slice(mid + 1))
  return root
}

Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution(object):

    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        if not nums:
            return None
        mid = len(nums) // 2
        root = TreeNode(nums[mid])
        root.left = self.sortedArrayToBST(nums[0:mid])
        root.right = self.sortedArrayToBST(nums[mid + 1:])
        return root

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func sortedArrayToBST(nums []int) *TreeNode {
	if len(nums) == 0 {
		return nil
	}

	length, mid := len(nums), 0
	if length%2 == 0 {
		mid = length / 2
	} else {
		mid = (length - 1) / 2
	}

	root := &TreeNode{Val: nums[mid]}
	root.Left = sortedArrayToBST(nums[:mid])
	root.Right = sortedArrayToBST(nums[mid+1:])

	return root
}