Converting a sorted doubly linked list to a BST

问题

How can a sorted doubly linked list be converted to a balanced binary search tree.

I was thinking of doing this the same way as converting an array to a balanced BST. Find the centre and then recursively convert the left part and the right part of the DLL. For example,

1 2 3 4 5 => 1 2 (3) 4 5 =>

     3
   /   \
  2     4
 /       \
1         5

This is leads to the recurrence T(n) = 2T(n/2) + O(n). O(n) is for finding the centre. The time complexity is therefore O(nlogn). I was wondering if there is an algorithm that does this in O(n).

回答1:

Yes there is O(n) solution. Note that an in-order traversal on a BST, is iterating the elements in the desired order, so just do an inorder traversal on an initially empty tree of size n, and fill it with elements in the list. [The i'th element you insert to the tree in your traversal, is the i'th element in the list].
At the end of the answer I added how to create an empty balanced tree in O(n).

pseudocode: [assuming |list| == |tree|]

global current <- null
fillTree(tree,list):
  current <- list.head
  fillTree(tree)
fillTree(tree):
  if tree == null:
     return
  fillTree(tree.left)
  //in-order traversal: we set the value after setting left, and before calling right
  tree.val <- current.val
  current <- current.next
  fillTree(tree.right)

Complexity is trivially O(n), since there is excactly one iteration for each vertex of the tree, and each iteration is O(1).

EDIT:
You can create an empty balanced tree, by simply building an empty complete tree(*), it is balanced and building it is O(n).

(*)A complete binary tree is a binary tree in which every level, except possibly the last, is completely filled.

回答2:

almost 4 years late. but here comes my functional solution. following is my code in haskell, complexity is also O(n):

import Data.List hiding (insert)

data Color = R | B deriving Show
data RBTree a = RBEmpty | RBTree { color :: Color
                                 , ltree :: RBTree a
                                 , nod   :: a
                                 , rtree :: RBTree a } deriving Show

fromOrdList ::  Ord e => [e] -> RBTree e
fromOrdList [] = empty
fromOrdList lst = 
    let (res, _) = _fol lst $ length lst
    in res
    where _fol :: (Ord e, Integral a) => [e] -> a -> (RBTree e, Maybe (e, [e]))
          _fol l 0            = (empty, uncons l)
          _fol (h:l) 1        = (RBTree B empty h empty, uncons l)
          _fol (h1:h2:l) 2    = (RBTree B (RBTree R empty h1 empty) h2 empty, uncons l)
          _fol (h1:h2:h3:l) 3 = (RBTree B (RBTree R empty h1 empty) h2 (RBTree R empty h3 empty), uncons l)
          _fol l n            =
            let mid                  = n `div` 2
                (ltr, Just (rt, tl)) = _fol l mid
                (rtr, mayremain)     = _fol tl (n - 1 - mid)
in (RBTree B ltr rt rtr, mayremain)

which is actually a part of my personal practise: https://github.com/HuStmpHrrr/PFDSPractise/blob/master/src/Tree/RBTree.hs#L97

回答3:

Look at my implementation of recursive insertion (c#). There are T(n) = 2*T(n/2) + O(1). O(1) is for finding the centre: (l+r)/2. So complicity is O(n)

public class Tree<T>
  {
    public class TreeNode<T>
    {
      public TreeNode<T> Right { get; set; }
      public TreeNode<T> Left { get; set; }
      public T Data { get; set; }
    }

    public Tree()
    {
      Root = new TreeNode<T>();
    }  

    public TreeNode<T> Root { get; set; }

    private void InsertSortedListRec(IList<T> items, TreeNode<T> curNode, int l, int r)
    {
      var mid = (l + r)/2;
      curNode.Data = items[mid];

      if (mid - 1 >= l)
      {
        curNode.Left = new TreeNode<T>();
        InsertSortedListRec(items, curNode.Left, l, mid - 1);
      }

      if (mid + 1 <= r)
      {
        curNode.Right = new TreeNode<T>();
        InsertSortedListRec(items, curNode.Right, mid + 1, r);
      }
  }

    public void InsertSortedList(IList<T> items)
    {
      InsertSortedListRec(items, Root, 0, items.Count - 1);
    }
  }

I assume that we have indexed array (we can convert linked list to array O(n))

来源：https://stackoverflow.com/questions/7874517/converting-a-sorted-doubly-linked-list-to-a-bst

标签

algorithm

binary-tree

doubly-linked-list