leetcode 将已排序的 数组/链表 转换为二叉搜索树（BST）,Python实现

思路：不论是数组还是链表，递归地找到他的root（即序列的中点），并返回。

1. 将数组转换为二叉树：

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

class Solution:
    # @param num, a list of integers
    # @return a tree node
    # 12:37
    def sortedArrayToBST(self, num):
        if not num:
            return None
        
        mid = len(num)//2  #“//”表示整数除法；“/”浮点数除法；
        root = TreeNode(num[mid])
        left = num[:mid]
        right = num[mid+1:]
        root.left = self.sortedArrayToBST(left)
        root.right = self.sortedArrayToBST(right)
        return root

2. 将链表转换为二叉树：

# 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):
        if not head:
            return None
        if not head.next:
            return TreeNode(head.val)
            
        Lroot = self.find_root(head)
        Troot = TreeNode(Lroot.val)
        Troot.left = self.sortedListToBST(head)
        Troot.right = self.sortedListToBST(Lroot.next)
        return Troot
        
    def find_root(self,head):  #找到链表的中点作为后半部分的起点，并将链表切断。
        if not head or not head.next:
            return head
        
        slow = head
        fast = head
        pre = head
        while fast and fast.next:
            pre = slow
            slow = slow.next
            fast = fast.next.next
        
        pre.next = None
        return slow