108. 将有序数组转换为二叉搜索树
题意
将有序数组转化为链表；
解题思路
将数组切片，中间值作为根结点，两边则则作为左子树和右子树；
维护左右两个值，实现对数组切片的功能；
实现
class Solution(object):
def sortedArrayToBST(self, nums):
"""
:type nums: List[int]
:rtype: TreeNode
"""
if not nums:
return None
elif len(nums) == 1:
return TreeNode(nums[0])
mid = len(nums) // 2
cur = TreeNode(nums[mid])
cur.left = self.sortedArrayToBST(nums[:mid])
cur.right = self.sortedArrayToBST(nums[mid+1:])
return cur
def sortedArrayToBST(self, nums):
"""
:type nums: List[int]
:rtype: TreeNode
"""
def dfs(start, end):
if start == end:
return None
mid = start + (end - start) // 2
cur = TreeNode(nums[mid])
cur.left = dfs(start, mid)
cur.right = dfs(mid+1, end)
return cur
if not nums:
return None
return dfs(0, len(nums))
def sortedArrayToBST(self, nums):
"""
:type nums: List[int]
:rtype: TreeNode
"""
def helper(left, right):
if left > right:
return None
if left == right:
return TreeNode(nums[left])
mid = (left + right) // 2
node = TreeNode(nums[mid])
node.left = helper(left, mid-1)
node.right = helper(mid+1, right)
return node
if not nums:
return None
return helper(0, len(nums)-1)