题解 | 将升序数组转化为平衡二叉搜索树

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param nums int整型一维数组
     * @return TreeNode类
     */
    public TreeNode sortedArrayToBST (int[] nums) {

          TreeNode  root = null;

        if (nums == null || nums.length == 0) {
            return root;
        } 

        if (nums.length == 1) {

            return new TreeNode(nums[0]);
        }
        int i = nums.length / 2   ;

        root=new TreeNode(nums[i]);

        // 分割为左侧和右侧数组
        int[] leftPart = Arrays.copyOfRange(nums, 0, i );
        int[] rightPart = Arrays.copyOfRange(nums, i + 1, nums.length);
        root.left = sortedArrayToBST(leftPart);
        root.right = sortedArrayToBST(rightPart);
        // write code here
        return root;
    }
}

思路：二分法 切割 然后 递归即可