Convert Sorted Array to Binary Search Tree

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

题解完全抄自ref的说明，感谢！

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题解：

先复习下什么是二叉搜索树（引自Wikipedia）：

二叉查找树（Binary Search Tree），也称有序二叉树（ordered binary tree）,排序二叉树（sorted binary tree），是指一棵空树或者具有下列性质的二叉树：

1. 若任意节点的左子树不空，则左子树上所有结点的值均小于它的根结点的值；
2. 任意节点的右子树不空，则右子树上所有结点的值均大于它的根结点的值；
3. 任意节点的左、右子树也分别为二叉查找树。

 

 

再复习下什么是平衡二叉树（引自GeekforGeek）：

An empty tree is height-balanced. A non-empty binary tree T is balanced if:
   1) Left subtree of T is balanced
   2) Right subtree of T is balanced
   3) The difference between heights of left subtree and right subtree is not more than 1.  

 

解决方法是选中点构造根节点，然后递归的构造左子树和右子树。

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Tree 构建基本题 http://www.cnblogs.com/springfor/p/3879823.html

public class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if(nums==null || nums.length==0) return null;
        return helper(nums, 0, nums.length-1);
    }
    public TreeNode helper(int[] num, int low, int high){
        if(low>high) return null;
        int mid  = (low+high)/2;
        TreeNode root = new TreeNode(num[mid ]);
        root.left = helper(num,low, mid-1);
        root.right = helper(num, mid+1,high);
        return root;
    }
}