# 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;
}
};