108 - Convert Sorted Array To Binary Search Tree

#easy
Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree.

Example 1:

Input: nums = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: [0,-10,5,null,-3,null,9] is also accepted:

Example 2:

Input: nums = [1,3]
Output: [3,1]
Explanation: [1,null,3] and [3,1] are both height-balanced BSTs.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode * traversal( vector<int>& nums, int left, int right ) {
        if ( left > right ) return NULL;
        int mid = ( left + right ) / 2;
        TreeNode * root = new TreeNode( nums[mid] );
        root -> left = traversal( nums, left, mid - 1 );
        root -> right = traversal( nums, mid + 1, right );
        return root;
    }


    TreeNode* sortedArrayToBST( vector<int>& nums) {
        return traversal( nums, 0, nums.size() - 1 );
    }
};