leetcode第108题将有序数组转换为二叉搜索树

[树，二叉搜索树，递归，平衡二叉树]

方法一：递归分治
因为数组是排序好的，将数组从中间一分为二，取中间的树为根节点，左边的数组用于构建左子树，右边的数组用于构建右子树。因为两边节点数量一致，可以保证树高度平衡。
递归处理，直到数组元素用尽为止。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if(nums.length==0)return null;
        TreeNode root;
        root = createTree(nums,0,nums.length-1);
        return root;
    }
    
    public TreeNode createTree(int[]nums,int left,int right){
        int length = right -left;
        if(length<0)return null;
        else if(length ==0)return new TreeNode(nums[left]);
        else{
            int mid= (right-left)/2+left;
             TreeNode root = new TreeNode(nums[mid]); 
            if(mid-left>=1) {
                TreeNode leftTreeNode = createTree(nums,left,mid-1);
                if(leftTreeNode!=null) {
                root.left=leftTreeNode;
                }
            }
            if(right-mid>=1) {
                TreeNode rightTreeNode = createTree(nums,mid+1,right);
                if(null!=rightTreeNode)
                    root.right = rightTreeNode;
            }
        return root;
        }
    }
}