Convert Sorted Array to Binary Tree

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

Solution

# 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
        """
        return self.createBST(nums, 0, len(nums) - 1)

    def createBST(self, nums, start, end):
        if start > end:
            return None
        mid = (start + end) / 2
        root = TreeNode(nums[mid])
        root.left = self.createBST(nums, start, mid - 1)
        root.right = self.createBST(nums, mid + 1, end)
        return root

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

    public TreeNode helper(int [] nums, int start, int end) {
        if(start > end) return null;
        int mid = (start + end) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = helper(nums, start, mid - 1);
        root.right = helper(nums, mid + 1, end);
        return root;
    }
}