题目
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
题目大意
将链表转化为高度平衡的二叉搜索树。高度平衡的定义：每个结点的 2 个子结点的深度不能相差超过 1 。
解题思路
思路比较简单，依次把链表的中间点作为根结点，类似二分的思想，递归排列所有结点即可。
参考代码
package leetcode
import (
"github.com/halfrost/LeetCode-Go/structures"
)
// ListNode define
type ListNode = structures.ListNode
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
// TreeNode define
type TreeNode = structures.TreeNode
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func sortedListToBST(head *ListNode) *TreeNode {
if head == nil {
return nil
}
if head != nil && head.Next == nil {
return &TreeNode{Val: head.Val, Left: nil, Right: nil}
}
middleNode, preNode := middleNodeAndPreNode(head)
if middleNode == nil {
return nil
}
if preNode != nil {
preNode.Next = nil
}
if middleNode == head {
head = nil
}
return &TreeNode{Val: middleNode.Val, Left: sortedListToBST(head), Right: sortedListToBST(middleNode.Next)}
}
func middleNodeAndPreNode(head *ListNode) (middle *ListNode, pre *ListNode) {
if head == nil || head.Next == nil {
return nil, head
}
p1 := head
p2 := head
for p2.Next != nil && p2.Next.Next != nil {
pre = p1
p1 = p1.Next
p2 = p2.Next.Next
}
return p1, pre
}
