How to merge two BST's efficiently?

问题:

How to merge two binary search trees maintaining the property of BST?

If we decide to take each element from a tree and insert it into the other, the complexity of this method would be O(n1 * log(n2)), where n1 is the number of nodes of the tree (say T1), which we have splitted, and n2 is the number of nodes of the other tree (say T2). After this operation only one BST has n1 + n2 nodes.

My question is: can we do any better than O(n1 * log(n2))?

回答1:

Naaff's answer with a little more details:

• Flattening a BST into a sorted list is O(N)
  • It's just "in-order" iteration on the whole tree.
  • Doing it for both is O(n1+n2)
• Merging two sorted lists is into one sorted list is O(n1+n2).
  • Keep pointers to the heads of both lists
  • Pick the smaller head and advance its pointer
  • This is how the merge of merge-sort works
• Creating a perfectly balanced BST from a sorted list is O(N)
  • See code snippet below for algorithm[1]
  • In our case the sorted list is of size n1+n2. so O(n1+n2)
  • The resulting tree would be the conceptual BST of binary searching the list

Three steps of O(n1+n2) result in O(n1+n2)

For n1 and n2 of the same order of magnitude, that's better than O(n1 * log(n2))

[1] Algorithm for creating a balanced BST from a sorted list (in Python):

def create_balanced_search_tree(iterator, n):
    if n == 0:
        return None
    n_left = n//2
    n_right = n - 1 - n_left
    left = create_balanced_search_tree(iterator, n_left)
    node = iterator.next()
    right = create_balanced_search_tree(iterator, n_right)
    return {'left': left, 'node': node, 'right': right}

回答2:

• Flatten trees into sorted lists.
• Merge sorted lists.
• Create tree out of merged list.

IIRC, that is O(n1+n2).

回答3:

What about flattening both trees into sorted lists, merging the lists and then creating a new tree?

回答4:

Jonathan,

After sorting, we have a list of length n1+n2. Building a binary tree out of it will take log(n1+n2) time. This is same as merge sort, only that at each recursive step we wont have a O(n1+n2) term as we have in merge sort algorithm . So the time complexity is log(n1+n2).

Now the complexity of the whole problem is O(n1+n2).

Also I would say this approach is good if two lists are comparable size. If the sizes are not comparable then it is best to insert every node of the small tree into a large tree. This would take O(n1*log(n2)) time. For example if we have two trees one of size 10 and another of size 1024. Here n1+n2 = 1034 where as n1log(n2) = 10*10 = 100. So approach has to depend upon the sizes of the two trees.

回答5:

O(n1 * log(n2)) is the average case scenario even if we have 2 merge any unsorted list into a BST. We are not utilizing the fact that list is sorted list or a BST.

According to me Lets assume one BST has n1 elements and other has n2 elements. Now convert one BST into a Sorted Array List L1 in O(n1).

Merged BST(BST, Array)

if (Array.size == 0) return BST if(Array.size ==1) insert the element in the BST. return BST;

Find the index in the array whose left element < BST.rootnode and right element >=BST.rootnode say Index. if(BST.rootNode.leftNode ==null ) //i.e No left node { insert all the array from Index to 0 into left of BST and } else { Merged BST(BST.leftNode, Array{0 to Index}) }

if(BST.rootNode.rightNode ==null)//i.e No right node { insert all the array from Index to Array.size into right of BST } else { Merged BST(BST.rightNode, Array{Index to Array.size}) }

return BST.

This algorithm will take << time than O(n1 * log(n2)) as every time we are partitioning the array and BST to handle the subproblem.

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回答6:

The idea is to use iterative inorder traversal. We use two auxiliary stacks for two BSTs. Since we need to print the elements in sorted form, whenever we get a smaller element from any of the trees, we print it. If the element is greater, then we push it back to stack for the next iteration.