LeetCode OJ：Convert Sorted Array to Binary Search Tree（将排序好的数组转换成二叉搜索树）

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

讲一个排序好的数组转换成二叉搜索树，这题没想出来，基本上是参考别人的，边界条件应该注意一下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return createTree(nums, 0, nums.size() - 1);
    }

    TreeNode * createTree(vector<int> & nums, int left, int right)
    {
        if(left > right)
            return NULL;
        int mid = left + (right - left)/2;
        TreeNode * leftNode = createTree(nums, left, mid - 1);
        TreeNode * rightNode = createTree(nums, mid + 1, right);
        TreeNode * tmpRoot=  new TreeNode(nums[mid]);
        tmpRoot->left = leftNode;
        tmpRoot->right = rightNode;
        return tmpRoot;
    }
};

 java版本的代码如下所示，看到排好序的其实就应该注意一点了，自己肯定是不需要再进行比较插入的：

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