[Swift]LeetCode108. 将有序数组转换为二叉搜索树 | Convert Sorted Array to Binary Search Tree

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Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / 
   -3   9
   /   /
 -10  5

---

将一个按照升序排列的有序数组，转换为一棵高度平衡二叉搜索树。

本题中，一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。

示例:

给定有序数组: [-10,-3,0,5,9],

一个可能的答案是：[0,-3,9,-10,null,5]，它可以表示下面这个高度平衡二叉搜索树：

      0
     / 
   -3   9
   /   /
 -10  5

---

36ms

/**
 * 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 sortedArrayToBST(_ nums: [Int]) -> TreeNode? {
        return makeTree(nums, 0, nums.count - 1)
    }
    
    func makeTree(_ nums: [Int], _ start: Int, _ end: Int) -> TreeNode? {
        guard start <= end else { return nil }

        let mid = (start + end) / 2
        let node = TreeNode(nums[mid])
        node.left = makeTree(nums, start, mid - 1)
        node.right = makeTree(nums, mid + 1, end)
        return node
    }
}

---

40ms

/**
 * 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 sortedArrayToBST(_ nums: [Int]) -> TreeNode? {
        guard nums.count != 0 else {
            return nil
        }
        
        let root = TreeNode(nums[nums.count / 2])
        root.left = sortedArrayToBST(Array(nums[0..<nums.count / 2]))
        root.right = sortedArrayToBST(Array(nums[nums.count / 2 + 1..<nums.count]))

        return root
    }
}

---

40ms

/**
 * 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 sortedArrayToBST(_ nums: [Int]) -> TreeNode? {
        guard nums.count > 0 else {
            return nil
        }

        guard nums.count > 1 else {
            let treeNode = TreeNode(nums[0])
            return treeNode
        }

        let middleIndex = nums.count/2
        let treeNode = TreeNode(nums[middleIndex])

        
        let leftSubarray = Array(nums[0..<middleIndex])
        let rightSubarray = ((middleIndex+1) < nums.count)
                            ? Array(nums[(middleIndex+1)..<nums.count])
                            : []

        let leftNode = sortedArrayToBST(leftSubarray)
        let rightNode = sortedArrayToBST(rightSubarray)

        treeNode.left = leftNode
        treeNode.right = rightNode
        return treeNode
    }
}

---

44ms

/**
 * 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 sortedArrayToBST(_ nums: [Int]) -> TreeNode? {
    return sortedArrayToBST1(nums, 0, nums.count)
  }
  
  private func sortedArrayToBST1(_ nums: [Int], _ beg: Int, _ end: Int) -> TreeNode? {
    if beg == end { return nil }
    if end - beg == 1 { return TreeNode(nums[beg]) }
    
    let mid = (beg + end) / 2
    let node = TreeNode(nums[mid])
    node.left = sortedArrayToBST1(nums, beg, mid)
    node.right = sortedArrayToBST1(nums, mid + 1, end)
    return node
  }
}

---

52ms

/**
 * 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 sortedArrayToBST(_ nums: [Int]) -> TreeNode? {
        if nums.count == 0 {
            return nil
        }
        if nums.count == 1 {
            return TreeNode(nums[0])
        }
        
        let mid = nums.count / 2
        let root = TreeNode(nums[mid])
        
        root.left = sortedArrayToBST(Array(nums[0..<mid]))
        if mid + 1 < nums.count {
            root.right = sortedArrayToBST(Array(nums[mid+1..<nums.count]))
        }
        return root
    }
}

---

1052ms

/**
 * 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 sortedArrayToBST(_ nums: [Int]) -> TreeNode? {
         if nums.count == 0 {
            return nil;
        }
        
       
        
        return createTree(nums, 0, nums.count - 1)
    }
       func createTree(_ nums: [Int], _ left: Int, _ right: Int) -> TreeNode? {
        
        if left > right {
            return nil
        }
        
        let mid = (left + right + 1) / 2 
        let node = TreeNode(nums[mid])
        
        node.left = createTree(nums, left, mid - 1)
        node.right = createTree(nums, mid + 1, right)
        
        return node;
    }
}