算法：有序数组转为平衡的二叉搜索树

Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

二叉搜索树的特点就是节点的左儿子所有点的值都小于该节点，其右儿子所有点的值都大于该节点。既然数组是有序的，先取数组的中间节点为根，然后把数组分为左右两个字数组，依次递归就可以了。上代码：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedArrayToBST(int[] num) {
        if(num==null||num.length==0)
        {
            return null;
        }
        int len=num.length;
        int mid=len/2;
        TreeNode root=new TreeNode(num[mid]);
        bst(root,num,0,mid-1,true);
        bst(root,num,mid+1,len-1,false);
        return root;
        
    }
    public void bst(TreeNode root,int[] a,int s,int e,boolean isLeft)
    {
        if(s>e)
        {
            return;
        }
        int mid=(s+e)/2;
        TreeNode node=new TreeNode(a[mid]);
        if(isLeft)
        {
            root.left=node;
            bst(node,a,s,mid-1,true);
            bst(node,a,mid+1,e,false);
        }
        else
        {
            root.right=node;
            bst(node,a,s,mid-1,true);
            bst(node,a,mid+1,e,false);
        }
    }
}