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.

A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.

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.

Constraints:

• 1 <= nums.length <= 104

• -104 <= nums[i] <= 104

• nums is sorted in a strictly increasing order.

解題

這題還滿經典的。

Runtime: 0 ms, faster than 100.00%

Memory Usage: 3.5 MB, less than 76.66%

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func sortedArrayToBST(nums []int) *TreeNode {
    var create func(int, int) *TreeNode
    create = func(start int, end int) *TreeNode {
        if start > end { return nil }
    
        mid := (start + end)/2

        root := &TreeNode{ Val: nums[mid] }

        root.Left = create(start, mid-1)
        root.Right = create(mid+1, end)
        
        return root
    }
    
    return create(0, len(nums)-1)
}

討論區看到的厲害解法

func sortedArrayToBST(nums []int) *TreeNode {
    if len(nums) == 0 {
        return nil
    }
    return &TreeNode{
        Val: nums[len(nums)/2],
        Left: sortedArrayToBST(nums[:len(nums)/2]),
        Right: sortedArrayToBST(nums[len(nums)/2 + 1:]),
    }
}