从有序链表创造平衡二叉搜索树 [英] Create Balanced Binary Search Tree from Sorted linked list

问题描述

什么是从排序单链表建立一个平衡的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

如何创建节点自下而上的？

How about creating nodes bottom-up?

此解决方案的时间复杂度为O（N）。详细解释在我的博客文章：

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

的http：// www.leet code.com / 2010/11 /转换排序列表到平衡binary.html

二遍历链表是我们所需要的。优先遍历获得列表的长度（其然后被传递作为参数n到函数），然后创建根据列表的顺序节点

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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