力扣hot100 - 108、将有序数组转换为二叉搜索树

题目：思路：将数组从最中间分为中间节点，左边数组和右边数组，中间作为根节点，左边作为左子树，右边作为右子树，左子树右子树同样取中间节点作为根节点。

/**
 * 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 build(nums , 0 , nums.length - 1);
    }

    TreeNode build(int [] nums , int left ,int right){
        if( left > right) return null;
        int mid = (left + right)/2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = build(nums , left , mid - 1);
        root.right = build(nums , mid + 1 , right);
        return root;
    }
}