题目

109. Convert Sorted List to Binary Search Tree

Given the head of 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 1:

Input: head = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]

Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.
Example 2:

Input: head = []
Output: []

Example 3:

Input: head = [0]
Output: [0]

Example 4:

Input: head = [1,3]
Output: [3,1]

Constraints:

The number of nodes in head is in the range [0, 2 * 10^4].
-10^5 <= Node.val <= 10^5

解法一：转为数组后，根据数组序号二分组装成二叉排序树

思路：
因为树有左右分叉，用递归解法。

把链表的值，组装成数组；
根据数组的索引，递归调用组装成二叉排序树

# Definition for singly-linked list.
class ListNode:def __init__(self, val=0, next=None):self.val = valself.next = next
# Definition for a binary tree node.
class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = right
class ConvertSortedListToBinarySearchTree:# Convert the given linked list to an arrydef mapListToValues(self, head: ListNode):vals = []while head:vals.append(head.val)head = head.nextreturn valsdef sortedListToBST(self, head: ListNode) -> TreeNode:# check edgeif head == None :return Noneif head.next == None:return TreeNode(head.val)# Form an array out of the given linked list and then# use the array to form the BSTvalues = self.mapListToValues(head)# l and r represent the start and end of the given arraydef convertListToBST(l, r):# Invalid caseif l > r:return None# Middle element forms the root.mid = (l + r) // 2node = TreeNode(values[mid])# Baes case for when there is only one element left in the arrayif l == r:return node# Recursively fom BST on the two halvesnode.left = convertListToBST(l, mid - 1)node.right = convertListToBST(mid + 1, r)return nodereturn convertListToBST(0, len(values) - 1)

解法二：找链表中的中间节点，组装成二叉排序树

组装树的方法是用递归，如果能持续找到中间节点，就可以持续组装树的middle, middle.left, middle.right.
找中间节点的方法，用快慢两个指针，慢指针每次走一步，快指针每次走两步；这样子就可以找出中间节点。

# Definition for singly-linked list.
class ListNode:def __init__(self, val=0, next=None):self.val = valself.next = next
# Definition for a binary tree node.
class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = right
class ConvertSortedListToBinarySearchTree:def findMiddle(self, head: ListNode):# The pointer used to disconnect the left half from the mid node.preNode = NoneslowNode = headfastNode = head# Iterate until fastNode doesn't reach the end of the linked list.while fastNode and fastNode.next:preNode = slowNodeslowNode = slowNode.nextfastNode = fastNode.next.next# Handling the case when slowNode was equal to headif preNode:preNode.next = Nonereturn slowNodedef sortedListToBST2(self, head: ListNode):# If the head doesn't exist, then the linked list is emptyif not head:return None# Find the middle element for the listmid = self.findMiddle(head)# the mid becones the root of the BST.node = TreeNode(mid.val)# Base case when there is just one element in the linked listif head == mid:return node# Recursively form balanced BSTs using the left and right halves of the original list.node.left = self.sortedListToBST2(head)node.right = self.sortedListToBST2(mid.next)return node

参考

https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/solution/

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算法：把排好序的链表转换为二叉排序树Convert Sorted List to Binary Search Tree

题目

解法一：转为数组后，根据数组序号二分组装成二叉排序树

解法二：找链表中的中间节点，组装成二叉排序树

参考

算法：把排好序的链表转换为二叉排序树Convert Sorted List to Binary Search Tree相关推荐

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