【Convert Sorted Array to Binary Search Tree】cpp

题目：

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

代码：

/**
 * 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) {
            const int len = nums.size();
            return Solution::sortedArr2BST(nums, , len-);
    }
    static TreeNode* sortedArr2BST(vector<int>& nums, int begin, int end )
    {
            if ( begin>end ) return NULL;
            int mid = (begin+end+)/;
            TreeNode *root = new TreeNode(nums[mid]);
            root->left = Solution::sortedArr2BST(nums, begin, mid-);
            root->right = Solution::sortedArr2BST(nums, mid+, end);
            return root;
    }
};

tips：

利用二分查找+递归。

注意终止条件：begin>end，返回NULL。

===================================

第二次过着这道题，代码与第一次写的几乎一样。

/**
 * 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 Solution::transform(nums, , nums.size()-);
        }
        static TreeNode* transform(vector<int>& nums, int begin, int end)
        {
            if ( begin>end ) return NULL;
            int mid = (begin+end)/;
            TreeNode* root = new TreeNode(nums[mid]);
            root->left = Solution::transform(nums, begin, mid-);
            root->right = Solution::transform(nums, mid+, end);
            return root;
        }
};