# 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

![](https://901207480-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LGKoChvN9am4__HCIRK%2F-LOSWpDeKKzRSEn0doCN%2F-LOSi64xkV33E2pxQi1b%2F%E8%9E%A2%E5%B9%95%E5%BF%AB%E7%85%A7%202018-10-10%20%E4%B8%8B%E5%8D%887.23.59.png?alt=media\&token=1d9e02af-361a-4720-b488-e12828186175)

## 2.想法 <a href="#id-2-xiang-fa" id="id-2-xiang-fa"></a>

* 提問
  * 確認題意:  height balanced
* function header, parameter input&#x20;
* 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** <a href="#id-3-cheng-shi" id="id-3-cheng-shi"></a>

```
/**
 * 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) {
        if (nums.size() == 0) {
            return NULL;
        }
        return buildTree(nums, 0, nums.size() - 1);
    }
private:
    TreeNode* buildTree(vector<int>& 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;
    }
};
```