0108. Convert Sorted Array to Binary Search Tree.md

BST

/**
 * 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* sortedArrayToBST(vector<int>& nums) {
        function<TreeNode*(int,int)> build = [&](int l, int r) -> TreeNode* {
            if (l > r) return nullptr;

            int mid = l + (r - l) / 2;
            TreeNode* root = new TreeNode(nums[mid]);

            root->left = build(l, mid - 1);
            root->right = build(mid + 1, r);

            return root;
        };

        return build(0, nums.size() - 1);
    }
};

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {number[]} nums
 * @return {TreeNode}
 */
var sortedArrayToBST = function(nums) {
    if(!nums.length) return null;
    
    let mid = Math.floor(nums.length / 2);
    let root = new TreeNode(nums[mid]);
    
    root.left = sortedArrayToBST(nums.slice(0, mid));
    root.right = sortedArrayToBST(nums.slice(mid + 1));

    return root;
};

Time Complexity: O(n)