将有序数组转换为二叉搜索树

/**
 * 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 {

    /**
    思路： 
        二分法找中点 作为根节点，保证左右两侧的元素个数相等，不导致树的倾斜
        递归创建树
     *//
    public TreeNode sortedArrayToBST(int[] nums) {
        return generate(nums, 0 ,nums.length-1);
    }
    public TreeNode generate(int[] nums , int start , int end ){
        if(start > end ){
            return null ;
        }
        int mid = (start + end) / 2 ;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = generate(nums,start , mid -1 ) ; 
        root.right = generate(nums,mid+1  , end);
        return root;
    }
}