从排序链表创建平衡二叉搜索树 [英] Create Balanced Binary Search Tree from Sorted linked list

问题描述

从排序的单向链表创建平衡二叉搜索树的最佳方法是什么?

What's the best way to create a balanced binary search tree from a sorted singly linked list?

推荐答案

如何自下而上创建节点?

How about creating nodes bottom-up?

这个解决方案的时间复杂度是 O(N).我的博文中有详细解释:

This solution's time complexity is O(N). Detailed explanation in my blog post:

http://www.leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html

我们只需要两次遍历链表.先遍历得到链表的长度(然后作为参数n传入函数)，然后按照链表的顺序创建节点.

Two traversal of the linked list is all we need. First traversal to get the length of the list (which is then passed in as the parameter n into the function), then create nodes by the list's order.

BinaryTree* sortedListToBST(ListNode *& list, int start, int end) {
  if (start > end) return NULL;
  // same as (start+end)/2, avoids overflow
  int mid = start + (end - start) / 2;
  BinaryTree *leftChild = sortedListToBST(list, start, mid-1);
  BinaryTree *parent = new BinaryTree(list->data);
  parent->left = leftChild;
  list = list->next;
  parent->right = sortedListToBST(list, mid+1, end);
  return parent;
}

BinaryTree* sortedListToBST(ListNode *head, int n) {
  return sortedListToBST(head, 0, n-1);
}

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