Merge pull request #25 from remy727/convert-sorted-list-to-binary-search-tree

109. Convert Sorted List to Binary Search Tree

Author: remy727

README.md

@@ -29,6 +29,7 @@
 | 91 | Decode Ways | [Ruby](./algorithms/ruby/0091-decode-ways.rb) | Medium |
 | 103 | Binary Tree Zigzag Level Order Traversal | [Ruby](./algorithms/ruby/0103-binary-tree-zigzag-level-order-traversal.rb) | Medium |
 | 104 | Maximum Depth of Binary Tree | [Ruby](./algorithms/ruby/0104-maximum-depth-of-binary-tree.rb) [Python3](./algorithms/python3/0104-maximum-depth-of-binary-tree.py) | Easy |
+| 109 | Convert Sorted List to Binary Search Tree | [Ruby](./algorithms/ruby/0109-convert-sorted-list-to-binary-search-tree.rb) | Medium |
 | 113 | Path Sum II | [Ruby](./algorithms/ruby/0113-path-sum-ii.rb) | Medium |
 | 121 | Best Time to Buy and Sell Stock | [Ruby](./algorithms/ruby/0121-best-time-to-buy-and-sell-stock.rb) | Easy |
 | 142 | Linked List Cycle II | [Ruby](./algorithms/ruby/0142-linked-list-cycle-ii.rb) | Medium |

algorithms/ruby/0109-convert-sorted-list-to-binary-search-tree.rb

@@ -0,0 +1,49 @@
+# frozen_string_literal: true
+
+# Problem: 109. Convert Sorted List to Binary Search Tree
+# URL: https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree
+
+=begin
+
+Given the head of a singly linked list where elements are sorted in ascending order, convert it to a
+height-balanced binary search tree.
+
+# Constraints:
+
+* The number of nodes in head is in the range [0, 2 * 104].
+* -105 <= Node.val <= 105
+
+=end
+
+# Definition for singly-linked list.
+# class ListNode
+# attr_accessor :val, :next
+# def initialize(val = 0, _next = nil)
+# @val = val
+# @next = _next
+# end
+# end
+# Definition for a binary tree node.
+# class TreeNode
+# attr_accessor :val, :left, :right
+# def initialize(val = 0, left = nil, right = nil)
+# @val = val
+# @left = left
+# @right = right
+# end
+# end
+# @param {ListNode} head
+# @return {TreeNode}
+def sorted_list_to_bst(head)
+ a = []
+ while head
+ a << head.val
+ head = head.next
+ end
+ f = -> i, j do
+ return nil if i >= j
+ m = (i + j) / 2
+ TreeNode.new a[m], f.(i, m), f.(m + 1, j)
+ end
+ f.(0, a.size)
+end