题解 | #将升序数组转化为平衡二叉搜索树# 分治法

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
     */
    TreeNode buildTree(int arr[]){
        int len = arr.length;
        if(len == 0){
            return null;
        }
        if(len == 1){
            return new TreeNode(arr[0]);
        }
        int midIdx = len/2;
        int midVal = arr[midIdx];
        TreeNode node = new TreeNode(midVal);
        node.left = buildTree(Arrays.copyOfRange(arr,0,midIdx));
        node.right = buildTree(Arrays.copyOfRange(arr,midIdx+1,len));
        return node;

    }
    public TreeNode sortedArrayToBST (int[] nums) {
        return buildTree(nums);
    }
}

取中间点，左右递归