108 Convert Sorted Array to Binary Search¶

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: Recursion¶

Recursively build a BST for a given range.

def build(nums, l, r): if l > r: return None m = l + (r – l) / 2 root = TreeNode(nums[m]) root.left = build(nums, l, m – 1) root.right = build(nums, m + 1, r) return root

return build(nums, 0, len(nums) – 1)

Time complexity: O(n) Space complexity: O(logn)

class Solution {
  public TreeNode sortedArrayToBST(int[] nums) {
    return buildBST(nums, 0, nums.length - 1);
  }

  private TreeNode buildBST(int[] nums, int l, int r) {
    if (l > r) return null;
    int m = l + (r - l) / 2;
    TreeNode root = new TreeNode(nums[m]);
    root.left = buildBST(nums, l, m - 1);
    root.right = buildBST(nums, m + 1, r);
    return root;
  }
}