标签：nts 有意思 null tree slow obs example represent while
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
class Solution { public TreeNode sortedListToBST(ListNode head) { List<Integer> list = new ArrayList(); while(head != null){ list.add(head.val); head = head.next; } if(list.size() == 0) return null; return help(list, 0, list.size() - 1); } public TreeNode help(List<Integer> list, int start, int end){ if(start > end) return null; int rootind = (end + start) / 2; TreeNode root = new TreeNode(list.get(rootind)); root.left = help(list, start, rootind - 1); root.right = help(list, rootind + 1, end); return root; } }
一定是end+start/2，不是减。
既然linkedlist无法随机访问，就把它先转化成list或者array，就和108一样。
public TreeNode sortedListToBST(ListNode head) { return sortedArrayToBST(head, null); } private TreeNode sortedArrayToBST(ListNode head, ListNode tail) { if (head == tail) { return null; } ListNode fast = head; ListNode slow = head; while (fast != tail && fast.next != tail) { slow = slow.next; fast = fast.next.next; } TreeNode root = new TreeNode(slow.val); root.left = sortedArrayToBST(head, slow); root.right = sortedArrayToBST(slow.next, tail); return root; }
这个快慢指针做法也很有意思
109. Convert Sorted List to Binary Search Tree
标签：nts 有意思 null tree slow obs example represent while
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/11444486.html
