# 【LeetCode】 108. Convert Sorted Array to Binary Search Tree

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

> For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

> 給一個經過漸增排序後的陣列，將它轉換為高度平衡的二元搜索樹(BST)。

> 對於這個問題，一個高度平衡的二元樹被定義為所有節點的兩個子樹，高度相差都不大於1的二元樹。

## Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

## Solution

* 要建立一個平衡的BST並不難，我們只要確定每次建立節點時都挑選中間的元素，就可以讓樹不會往某一邊歪斜。
* 我的作法是將陣列的中間值設為該節點的值，然後用剩下陣列的左半邊建立左子樹，右半邊建立右子樹。

### Code

/**
 * 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:
    void BST(vector<int>& nums, int s, int e, TreeNode*& n)
    {
        if(s > e) return;
        int mid = (s + e) / 2;
        n = new TreeNode(nums[mid]);
        BST(nums, s, mid - 1, n->left);
        BST(nums, mid + 1, e, n->right);
    }
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        TreeNode* root = NULL;
        BST(nums, 0, nums.size() - 1, root);
        return root;
    }
};

###### tags: `LeetCode` `C++`