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
难度：medium
题目：给定一个单链表其元素为升序排列，将其转换成高度平衡的二叉搜索树
思路：中序遍历
Runtime: 1 ms, faster than 99.17% of Java online submissions for Convert Sorted List to Binary Search Tree.Memory Usage: 41 MB, less than 9.56% of Java online submissions for Convert Sorted List to Binary Search Tree.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {val = x;}
* }
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) {val = x;}
* }
*/
class Solution {
public TreeNode sortedListToBST(ListNode head) {
if (null == head) {
return null;
}
ListNode ptr = head;
int count = 0;
for (; ptr != null; ptr = ptr.next, count++);
ListNode[] headList = {head};
return sortedListToBST(headList, 0, count – 1);
}
public TreeNode sortedListToBST(ListNode[] head, int start, int end) {
if (start > end) {
return null;
}
int mid = start + (end – start) / 2;
TreeNode left = sortedListToBST(head, start, mid – 1);
TreeNode root = new TreeNode(head[0].val);
root.left = left;
head[0] = head[0].next;
root.right = sortedListToBST(head, mid + 1, end);
return root;
}
}