# 108. Convert Sorted Array to Binary Search Tree
## 1.問題
* 利用給予的序列建構出balanced binary search tree
## 2.想法
* 提問
* 確認題意: height balanced
* function header, parameter input
* 說明想法
* BST定義為left descents <= n < right descents, 如果用二分法, 將每次的mid作為新的root node, 不僅左右高度相同, 而且滿足BST的定義
* 測試計算複雜度
## **3.程式碼**
```
/**
* 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& nums) {
if (nums.size() == 0) {
return NULL;
}
return buildTree(nums, 0, nums.size() - 1);
}
private:
TreeNode* buildTree(vector& nums, int left, int right) {
if (left > right) {
return NULL;
}
int mid = (left + right) / 2;
TreeNode* node = new TreeNode(nums[mid]);
node->left = buildTree(nums, left, mid - 1);
node->right = buildTree(nums, mid + 1, right);
return node;
}
};
```
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