108. Convert Sorted Array to Binary Search Tree
Read the full problem statement on LeetCode.
Difficulty: easy Acceptance: 74% Topics: Array, Divide and Conquer, Tree, Binary Search Tree, Binary Tree
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Reference solution (spoiler · python)
# Time: O(n)
# Space: O(logn)
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution(object):
def sortedArrayToBST(self, nums):
"""
:type nums: List[int]
:rtype: TreeNode
"""
return self.sortedArrayToBSTRecu(nums, 0, len(nums))
def sortedArrayToBSTRecu(self, nums, start, end):
if start == end:
return None
mid = start + self.perfect_tree_pivot(end - start)
node = TreeNode(nums[mid])
node.left = self.sortedArrayToBSTRecu(nums, start, mid)
node.right = self.sortedArrayToBSTRecu(nums, mid + 1, end)
return node
def perfect_tree_pivot(self, n):
"""
Find the point to partition n keys for a perfect binary search tree
"""
x = 1
# find a power of 2 <= n//2
# while x <= n//2: # this loop could probably be written more elegantly :)
# x *= 2
x = 1 << (n.bit_length() - 1) # use the left bit shift, same as multiplying x by 2**n-1
if x // 2 - 1 <= (n - x):
return x - 1 # case 1: the left subtree of the root is perfect and the right subtree has less nodes
else:
return n - x // 2 # case 2 == n - (x//2 - 1) - 1 : the left subtree of the root
# has more nodes and the right subtree is perfect.
# Time: O(n)
# Space: O(logn)
class Solution2(object):
def sortedArrayToBST(self, nums):
"""
:type nums: List[int]
:rtype: TreeNode
"""
self.iterator = iter(nums)
return self.helper(0, len(nums))
def helper(self, start, end):
if start == end:
return None
mid = (start + end) // 2
left = self.helper(start, mid)
current = TreeNode(next(self.iterator))
current.left = left
current.right = self.helper(mid+1, end)
return current
Solution from kamyu104/LeetCode-Solutions · MIT
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