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[Linked List] Add a solution to Convert Sorted List to Binary Search … #270
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Original file line number | Diff line number | Diff line change |
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/** | ||
* Question Link: https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/ | ||
* Primary idea: Convert the linked list to array, handle array using Divide and Conqure | ||
* Time Complexity Per Action: O(N), Space Complexity: O(N) | ||
* | ||
*/ | ||
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/** | ||
* Definition for singly-linked list. | ||
* public class ListNode { | ||
* public var val: Int | ||
* public var next: ListNode? | ||
* public init(_ val: Int) { | ||
* self.val = val | ||
* self.next = nil | ||
* } | ||
* } | ||
*/ | ||
/** | ||
* Definition for a binary tree node. | ||
* public class TreeNode { | ||
* public var val: Int | ||
* public var left: TreeNode? | ||
* public var right: TreeNode? | ||
* public init(_ val: Int) { | ||
* self.val = val | ||
* self.left = nil | ||
* self.right = nil | ||
* } | ||
* } | ||
*/ | ||
class Solution { | ||
func sortedListToBST(_ head: ListNode?) -> TreeNode? { | ||
var sortedArr: [Int] = [] | ||
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var cur = head | ||
while cur != nil { | ||
sortedArr.append(cur!.val) | ||
cur = cur!.next | ||
} | ||
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return _helper(sortedArr, 0, sortedArr.count - 1) | ||
} | ||
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private func _helper(_ sortedArr: [Int], _ left: Int, _ right: Int) -> TreeNode? { | ||
guard left <= right else { return nil } | ||
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let mid = left + (right - left) >> 1 | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. the |
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let root = TreeNode(sorted[mid]) | ||
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root.left = _helper(sortedArr, left, mid - 1) | ||
root.right = _helper(sortedArr, mid + 1, right) | ||
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return root | ||
} | ||
} |
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Do you think you can do something like this below, so you don't have to implicitly unwrap the variable which is not recommended in production code?