108. Convert Sorted Array to Binary Search Tree

Problem

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 (Java)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode sortedArrayToBST(int[] num) {
        /*1. Set up recursion*/
        return makeTree(num, 0, num.length - 1);
    }

    private TreeNode makeTree(int[] num, int left, int right) {
        /*2. left as lowest# can't be greater than right*/
        if (left > right) {
            return null;
        }
        /*3. Set median# as node*/
        int mid = (left + right) / 2;
        TreeNode midNode = new TreeNode(num[mid]);
        /*4. Set mid node's kids*/
        midNode.left = makeTree(num, left, mid - 1);
        midNode.right = makeTree(num, mid + 1, right);
        /*5. Sends node back || Goes back to prev node*/
        return midNode;
    }
}

Solution (Javascript)

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {number[]} nums
 * @return {TreeNode}
 */
var sortedArrayToBST = function(nums) {
  return helper(nums, 0, nums.length - 1);
};

var helper = function (nums, ll, rr) {
  if (ll > rr) return null;
  var mid = Math.ceil((ll + rr) / 2);
  var root = new TreeNode(nums[mid]);
  root.left = helper(nums, ll, mid - 1);
  root.right = helper(nums, mid + 1, rr);
  return root;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).