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108. Convert Sorted Array to Binary Search Tree

Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced* binary search tree*.

Example 1:

img

Input: nums = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: [0,-10,5,null,-3,null,9] is also accepted:

Example 2:

img

Input: nums = [1,3]
Output: [3,1]
Explanation: [1,null,3] and [3,1] are both height-balanced BSTs.

Constraints:

  • 1 <= nums.length <= 104
  • -104 <= nums[i] <= 104
  • nums is sorted in a strictly increasing order.

Solution:

/**
 * 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[] nums) {
       return dfs(nums, 0, nums.length - 1);
    }

    private TreeNode dfs(int[] nums, int left, int right){
        if (left > right){
            return null;
        }

        int mid = left + (right - left) /2;

        TreeNode root = new TreeNode(nums[mid]);

        root.left = dfs(nums, left, mid - 1);
        root.right = dfs(nums, mid + 1, right);

        return root;
    }
} 
// TC: O(n)
// SC: O(n)