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, , nums.size() - );
     }

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