For a sorted array, the left half will be in the left subtree, middle value as the root, right half in the right subtree. This holds through for every node:
[1, 2, 3, 4, 5, 6, 7] -> left:
[1, 2, 3], root:
[5, 6, 7]
[1, 2, 3] -> left:
[5, 6, 7] -> left:
Many of the approaches here suggest slicing an array recursively and passing them. However, slicing the array is expensive. It is better to pass the left and right bounds into recursive calls instead.
class Solution(object): def sortedArrayToBST(self, nums): # Time: O(n) # Space: O(n) in the case of skewed binary tree. def convert(left, right): if left > right: return None mid = (left + right) // 2 node = TreeNode(nums[mid]) node.left = convert(left, mid - 1) node.right = convert(mid + 1, right) return node return convert(0, len(nums) - 1)
@yangshun I agree with you that slice is expensive. Here is my solution similar to yours.
class Solution: def sortedArrayToBST(self, nums): """ :type nums: List[int] :rtype: TreeNode """ return self.build(nums, 0, len(nums)-1) def build(self, nums, low, high): if low == high: return TreeNode(nums[low]) elif low < high: mid = (low+high)//2 node = TreeNode(nums[mid]) node.left = self.build(nums, low, mid-1) node.right = self.build(nums, mid+1, high) return node