leetcode109. Convert Sorted List to Binary Search Tree

Posted 2021-11-23 seyjs

题目如下：

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

 

解题思路：题目没有明确要求不允许使用额外的内存，所以最简单的方法是把linked list中每个元素的值存入list，然后套用【leetcode】108. Convert Sorted Array to Binary Search Tree 的解法。

代码如下：

# 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 recursive(self,node,nums):
        mid = len(nums)/2
        left_num = nums[:mid]
        if len(left_num) > 0:
            node.left = TreeNode(left_num[len(left_num)/2])
            self.recursive(node.left,left_num)
        right_num = nums[mid+1:]
        if len(right_num) > 0:
            node.right = TreeNode(right_num[len(right_num)/2])
            self.recursive(node.right,right_num)
    def sortedListToBST(self, head):
        """
        :type head: ListNode
        :rtype: TreeNode
        """
        nums = []
        while head != None:
            nums.append(head.val)
            head = head.next
        if len(nums) == 0:
            return None
        root = TreeNode(nums[len(nums)/2])
        self.recursive(root,nums)
        return root