[LinkList/Tree]109. Convert Sorted List to Binary Search Tree

• 分类：LinkList/Tree
• 时间复杂度: O(nlogn) 因为每次都需要遍历linklist的一半
• 空间复杂度: O(h) 树的节点的深度

109. Convert Sorted List to Binary Search Tree

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

代码：

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

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

class Solution:
    def sortedListToBST(self, head: 'ListNode') -> 'TreeNode':
        # 没有或者只有一个数的时候
        if head==None:
            return None
        if head.next==None:
            return TreeNode(head.val)
        # 有两个数以上的时候
        slow,fast=head,head
        prev=None
        while fast and fast.next:
            prev=slow
            slow=slow.next
            fast=fast.next.next
        prev.next=None
        
        root=TreeNode(slow.val)
        root.left=self.sortedListToBST(head)
        root.right=self.sortedListToBST(slow.next)
        
        
        return root

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讨论：

1.一看之下感觉和108很像，后来发现是linklist，就又不会做了=。=于是在youtube上找到了公瑾讲解
2.这道题考察了linklist和树的操作
3.找linklist的中位数可以用fast/slow指针去找，每次fast走两格，slow走一格，当fast走完的时候slow刚好走到了中间
4.边界条件其实也很难，一定要注意，当没有数和只有一个数的时候是边界条件，直接退出即可，只有linklist有两个数+的时候才开始找中位数做题。