Convert Sorted List to Binary Search Tree

描述

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

分析

这题与上一题类似，但是单链表不能随机访问，而自顶向下的二分法必须需要RandomAccessIterator，因此前面的方法不适用本题。

存在一种自底向上(bottom-up)的方法，见 http://leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html

分治法，自顶向下

分治法，类似于 Convert Sorted Array to Binary Search Tree，自顶向下，复杂度 O(nlogn)。

  1. // Convert Sorted List to Binary Search Tree
  2. // 二分法，类似于 Convert Sorted Array to Binary Search Tree，
  3. // 自顶向下，时间复杂度O(nlogn)，空间复杂度O(logn)
  4. class Solution {
  5.     public TreeNode sortedListToBST (ListNode head) {
  6.         if(head == null) return null;
  7.         if(head.next == null) return new TreeNode(head.val);
  9.         ListNode mid = cutAtMiddle(head);
  11.         TreeNode root = new TreeNode(mid.val);
  12.         root.left = sortedListToBST(head);
  13.         root.right = sortedListToBST(mid.next);
  15.         return root;
  16.     }
  18.     ListNode cutAtMiddle(ListNode head) {
  19.         if(head == null) return null;
  21.         ListNode fast = head;
  22.         ListNode slow = head;
  23.         ListNode prev_slow = head;
  25.         while(fast != null && fast.next != null){
  26.             prev_slow = slow;
  27.             slow = slow.next;
  28.             fast = fast.next.next;
  29.         }
  31.         prev_slow.next = null;
  32.         return slow;
  33.     }
  34. }

自底向上

  1. // Convert Sorted List to Binary Search Tree
  2. // bottom-up，时间复杂度O(n)，空间复杂度O(logn)
  3. public class Solution {
  4.     public TreeNode sortedListToBST(ListNode head) {
  5.         int len = 0;
  6.         ListNode p = head;
  7.         while (p != null) {
  8.             len++;
  9.             p = p.next;
  10.         }
  11.         return sortedListToBST(new Container(head), 0, len - 1);
  12.     }
  13.     private static TreeNode sortedListToBST(Container list, int start, int end) {
  14.         if (start > end) return null;
  16.         int mid = start + (end - start) / 2;
  17.         TreeNode leftChild = sortedListToBST(list, start, mid - 1);
  18.         TreeNode parent = new TreeNode(list.p.val);
  19.         parent.left = leftChild;
  20.         list.p = list.p.next;
  21.         parent.right = sortedListToBST(list, mid + 1, end);
  22.         return parent;
  23.     }
  24.     // 模拟二级指针
  25.     static class Container {
  26.         ListNode p;
  27.         public Container(ListNode p) {
  28.             this.p = p;
  29.         }
  30.     }
  31. }

相关题目

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/binary-tree/bst/convert-sorted-list-to-binary-search-tree.html