Question description

Screenshot 2016-10-05 11.36.19.png

My code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        int length = nums.length;
        if (length < 1) return null;
        if (length == 1) {
            TreeNode root = new TreeNode(nums[0]);
            root.left = null;
            root.right = null;
            return root;
        }
        TreeNode root = new TreeNode(nums[length / 2]);
        
        int leftFrom = 0;
        int leftTo = length / 2;
        int rightFrom = length / 2 + 1;
        int rightTo = length;
        
        int[] leftNums = Arrays.copyOfRange(nums, leftFrom, leftTo);
        int[] rightNums = Arrays.copyOfRange(nums, rightFrom, rightTo);
        
        root.left = sortedArrayToBST(leftNums);
        root.right = sortedArrayToBST(rightNums);
        
        return root;
    }
}

Solution

A sorted array is a natural balanced binary search tree. Just find the middle number of nums and set it as root node, and do the same to the left sub-array and right sub-array.

Test result

Screenshot 2016-10-05 11.42.42.png