# 108. Convert Sorted Array to Binary Search Tree

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

# Solution

Approach 1: Recursion.

# Code (Python)

Approach 1:

    def sortedArrayToBST(self, nums: 'List[int]') -> 'TreeNode':
        return self._to_bst(nums, 0, len(nums) - 1)
    
    def _to_bst(self, nums, start, end):
        if start == end:
            return TreeNode(nums[start])
        if start > end:
            return None
        mid = (start + end) // 2
        root = TreeNode(nums[mid])
        root.left = self._to_bst(nums, start, mid - 1)
        root.right = self._to_bst(nums, mid + 1, end)
        return root

# Code (C++)

Approach 1:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return sortedArrayToBST(nums, 0, nums.size()-1);
    }
private:
    TreeNode* sortedArrayToBST(vector<int>& nums, int head, int tail) {
        if (head > tail) return NULL;
        int mid = head + (tail - head) / 2;
        TreeNode *node = new TreeNode(nums[mid]);
        node->left  = sortedArrayToBST(nums, head, mid-1);
        node->right = sortedArrayToBST(nums, mid+1, tail);
        return node;
    }
};