Convert Sorted List to Binary Search Tree

Description

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. Example: Given the sorted linked list: [-10,-3,0,5,9], One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

想法：

和108 Convert Sorted Array to Binary Search Tree 相似， 但是List無法直接透過index查找，概念一樣是從中間切一半，用快慢指針找到中間點，並要記錄slow前面一個node, 找到中點後以中點建立根節點root，接著把List斷開，左右兩邊的List都不包含中點，在對兩邊做遞迴 讓root連上左右兩邊 因為左邊部分做到最後會是slow == head，要另外設條件說當recursion做到這裡後，要停 止，右邊因為base case已經有 not head的狀況，所以不用額外再處理了

Code:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def sortedListToBST(self, head):
        """
        :type head: ListNode
        :rtype: TreeNode
        """
        if not head: return None
        if not head.next: return TreeNode(head.val)
        # 用快慢指針找到中點並切斷分為左右子樹
        prev = slow = fast = head
        while fast and fast.next:
            prev = slow
            slow = slow.next
            fast = fast.next.next
            
        second = slow.next
        prev.next = None
        current = TreeNode(slow.val)
        # if head != slow: # head == slow: 表示左邊沒了
        current.left = self.sortedListToBST(head)
        current.right = self.sortedListToBST(second)
        
        return current