108. 将有序数组转换为二叉搜索树

原题

https://leetcode.cn/problems/convert-sorted-array-to-binary-search-tree/description/

思路

递归

复杂度

时间：O(n)
空间：O(n)

Python代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
        if len(nums) == 0:
            return None
        mid = len(nums) // 2
        root = TreeNode(nums[mid])
        root.left = self.sortedArrayToBST(nums[:mid])
        root.right = self.sortedArrayToBST(nums[mid + 1 :])
        return root

Go代码

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func sortedArrayToBST(nums []int) *TreeNode {
	if len(nums) == 0 {
		return nil
	}
	mid := len(nums) / 2
	root := &TreeNode{Val: nums[mid]}
	root.Left = sortedArrayToBST(nums[:mid])
	root.Right = sortedArrayToBST(nums[mid+1:])
	return root
}