Merging 2 Binary Search Trees - algorithm

Question

How do you merge 2 Binary Search Trees in such a way that the resultant tree contains all the elements of both the trees and also maintains the BST property.
I saw the solution provided in
How to merge two BST's efficiently?
However that solution involves converting into a Double Linked List. I was wondering if there is a more elegant way of doing this which could be done in place without the conversion. I came up with the following pseudocode. Does it work for all cases? Also I am having trouble with the 3rd case.
node* merge(node* head1, node* head2) {
if (!head1)
return head2;
if (!head2)
return head1;
// Case 1.
if (head1->info > head2->info) {
node* temp = head2->right;
head2->right = NULL;
head1->left = merge(head1->left, head2);
head1 = merge(head1, temp);
return head1;
} else if (head1->info < head2->info) { // Case 2
// Similar to case 1.
} else { // Case 3
// ...
}
}

Answer 1

The two binary search trees (BST) cannot be merged directly during a recursive traversal.
Suppose we should merge Tree 1 and Tree 2 shown in the figure.
The recursion should reduce the merging to a simpler situation. We cannot reduce
the merging only to the respective left subtrees L1 and L2, because L2 can contain
numbers larger than 10, so we would need to include the right
subtree R1 into the process. But then we include numbers greater
than 10 and possibly greater than 20, so we would need to include
the right subtree R2 as well. A similar reasoning shows that
we cannot simplify the merging by including subtrees from Tree 1 and from Tree 2
at the same time.
The only possibility for reduction is to simplify only inside the respective trees.
So, we can transform
the trees to their right spines with sorted nodes:
Now, we can merge the two spines easily into one spine. This
spine is in fact a BST, so we could stop here. However, this BST
is completely unbalanced, so we transform it to a balanced BST.
The complexity is:
Spine 1: time = O(n1), space = O(1)
Spine 2: time = O(n2), space = O(1)
Merge: time = O(n1+n2), space = O(1)
Balance: time = O(n1+n2), space = O(1)
Total: time = O(n1+n2), space = O(1)
The complete running code is on http://ideone.com/RGBFQ. Here are the essential parts. The top level code is a follows:
Node* merge(Node* n1, Node* n2) {
Node *prev, *head1, *head2;
prev = head1 = 0; spine(n1, prev, head1);
prev = head2 = 0; spine(n2, prev, head2);
return balance(mergeSpines(head1, head2));
}
The auxiliary functions are for the tranformation to spines:
void spine(Node *p, Node *& prev, Node *& head) {
if (!p) return;
spine(p->left, prev, head);
if (prev) prev->right = p;
else head = p;
prev = p;
p->left = 0;
spine(p->right, prev, head);
}
Merging of the spines:
void advance(Node*& last, Node*& n) {
last->right = n;
last = n;
n = n->right;
}
Node* mergeSpines(Node* n1, Node* n2) {
Node head;
Node* last = &head;
while (n1 || n2) {
if (!n1) advance(last, n2);
else if (!n2) advance(last, n1);
else if (n1->info < n2->info) advance(last, n1);
else if (n1->info > n2->info) advance(last, n2);
else {
advance(last, n1);
printf("Duplicate key skipped %d \n", n2->info);
n2 = n2->right;
}
}
return head.right;
}
Balancing:
Node* balance(Node *& list, int start, int end) {
if (start > end) return NULL;
int mid = start + (end - start) / 2;
Node *leftChild = balance(list, start, mid-1);
Node *parent = list;
parent->left = leftChild;
list = list->right;
parent->right = balance(list, mid+1, end);
return parent;
}
Node* balance(Node *head) {
int size = 0;
for (Node* n = head; n; n = n->right) ++size;
return balance(head, 0, size-1);
}

Answer 2

Assuming we have two trees A and B we insert root of tree A into tree B and using rotations move inserted root to become new root of tree B. Next we recursively merge left and right sub-trees of trees A and B. This algorithm takes into account both trees structure but insertion still depends on how balanced target tree is. You can use this idea to merge the two trees in O(n+m) time and O(1) space.
The following implementation is due to Dzmitry Huba:
// Converts tree to sorted singly linked list and appends it
// to the head of the existing list and returns new head.
// Left pointers are used as next pointer to form singly
// linked list thus basically forming degenerate tree of
// single left oriented branch. Head of the list points
// to the node with greatest element.
static TreeNode<T> ToSortedList<T>(TreeNode<T> tree, TreeNode<T> head)
{
if (tree == null)
// Nothing to convert and append
return head;
// Do conversion using in order traversal
// Convert first left sub-tree and append it to
// existing list
head = ToSortedList(tree.Left, head);
// Append root to the list and use it as new head
tree.Left = head;
// Convert right sub-tree and append it to list
// already containing left sub-tree and root
return ToSortedList(tree.Right, tree);
}
// Merges two sorted singly linked lists into one and
// calculates the size of merged list. Merged list uses
// right pointers to form singly linked list thus forming
// degenerate tree of single right oriented branch.
// Head points to the node with smallest element.
static TreeNode<T> MergeAsSortedLists<T>(TreeNode<T> left, TreeNode<T> right, IComparer<T> comparer, out int size)
{
TreeNode<T> head = null;
size = 0;
// See merge phase of merge sort for linked lists
// with the only difference in that this implementations
// reverts the list during merge
while (left != null || right != null)
{
TreeNode<T> next;
if (left == null)
next = DetachAndAdvance(ref right);
else if (right == null)
next = DetachAndAdvance(ref left);
else
next = comparer.Compare(left.Value, right.Value) > 0
? DetachAndAdvance(ref left)
: DetachAndAdvance(ref right);
next.Right = head;
head = next;
size++;
}
return head;
}
static TreeNode<T> DetachAndAdvance<T>(ref TreeNode<T> node)
{
var tmp = node;
node = node.Left;
tmp.Left = null;
return tmp;
}
// Converts singly linked list into binary search tree
// advancing list head to next unused list node and
// returning created tree root
static TreeNode<T> ToBinarySearchTree<T>(ref TreeNode<T> head, int size)
{
if (size == 0)
// Zero sized list converts to null
return null;
TreeNode<T> root;
if (size == 1)
{
// Unit sized list converts to a node with
// left and right pointers set to null
root = head;
// Advance head to next node in list
head = head.Right;
// Left pointers were so only right needs to
// be nullified
root.Right = null;
return root;
}
var leftSize = size / 2;
var rightSize = size - leftSize - 1;
// Create left substree out of half of list nodes
var leftRoot = ToBinarySearchTree(ref head, leftSize);
// List head now points to the root of the subtree
// being created
root = head;
// Advance list head and the rest of the list will
// be used to create right subtree
head = head.Right;
// Link left subtree to the root
root.Left = leftRoot;
// Create right subtree and link it to the root
root.Right = ToBinarySearchTree(ref head, rightSize);
return root;
}
public static TreeNode<T> Merge<T>(TreeNode<T> left, TreeNode<T> right, IComparer<T> comparer)
{
Contract.Requires(comparer != null);
if (left == null || right == null)
return left ?? right;
// Convert both trees to sorted lists using original tree nodes
var leftList = ToSortedList(left, null);
var rightList = ToSortedList(right, null);
int size;
// Merge sorted lists and calculate merged list size
var list = MergeAsSortedLists(leftList, rightList, comparer, out size);
// Convert sorted list into optimal binary search tree
return ToBinarySearchTree(ref list, size);
}

Answer 3

The best way we could merge the trees in place is something like:
For each node n in first BST {
Go down the 2nd tree and find the appropriate place to insert n
Insert n there
}
Each iteration in the for loop is O(log n) since we are dealing with trees, and the for loop will be iterated n times, so in total we have O(n log n).

Answer 4

A BST is a ordered or sorted binary tree. My algorithm would be to simple :
traverse through both trees
compare the values
insert the smaller of the two into a new BST.
The python code for traversing is as follows:
def traverse_binary_tree(node, callback):
if node is None:
return
traverse_binary_tree(node.leftChild, callback)
callback(node.value)
traverse_binary_tree(node.rightChild, callback)
The cost for traversing through the BST and building a new merged BST would remain O(n)

Answer 5

This blog post provides a solution to the problem with O(logn) space complexity. (Pay attention that the given approach does not modify input trees.)

Answer 6

This can be done in 3 steps:
covert the BSTs to sorted linked list (this can be done in place with O(m+n) time)
Merge this two sorted linked lists to a single list (this can be done in place with O(m+n) time)
Convert sorted linked list to balanced BST (this can be done in place with O(m+n) time)

Answer 7

Here is what I would do.
This solution is O(n1+n2) time complexity.
STEPS:
Perform the inorder traversal of both the trees to get sorted arrays --> linear time
Merge the two arrays --> again linear time
Convert the merged array into a Balanced binary search tree --> again linear time
This would require O(n1+n2) time and space.
Links you may find useful while implementing:
How to merge 2 sorted arrays
Inorder traversal
Sorted array to a balanced BST

Answer 8

The following algorithm is from Algorithms in C++.
The idea is almost the same as in the algorithm posted by PengOne. This algorithm does in place merging, time complexity is O(n+m).
link join(link a, link b) {
if (b == 0) return a;
if (a == 0) return b;
insert(b, a->item);
b->left = join(a->left, b->left);
b->right = join(a->right, b->right);
delete a;
return b;
}
insert just inserts an item in the right place in the tree.
void insert(link &h, Item x) {
if (h == 0) {
h = new node(x);
return;
}
if (x.key() < h->item.key()) {
insert(h->left, x);
rotateRight(h);
}
else {
insert(h->right, x);
rotateLeft(h);
}
}
rotateRight and rotateLeft keep tree in the right order.
void rotateRight(link &h) {
link x = h->left;
h->left = x->right;
x->right = h;
h = x;
}
void rotateLeft(link &h) {
link x = h->right;
h->right = x->left;
x->left = h;
h = x;
}
Here link is node *.

Answer 9

Assuming the question is just to print sorted from both BSTs. Then the easier way is,
Store inorder traversal of 2 BSTs in 2 seperate arrays.
Now the problem reduces to merging\printing elements from 2 sorted arrays, which we got from step one. This merging can be done in o(m) when m>n or o(n) when m
Complexity: o(m+n)
Aux space: o(m+n) for the 2 arrays

Answer 10

MergeTwoBST_to_BalancedBST.java
public class MergeTwoBST_to_BalancedBST {
// arr1 and arr2 are the input arrays to be converted into a binary search
// structure and then merged and then balanced.
int[] arr1 = new int[] { 1, 2, 3, 4, 5, 6, 7, 8 };
int[] arr2 = new int[] { 11, 12, 13, 14, 15, 16, 17, 18 };
BSTNode root1;
BSTNode root2;
// vector object to hold the nodes from the merged unbalanced binary search
// tree.
Vector<BSTNode> vNodes = new Vector<BSTNode>();
/**
* Constructor to initialize the Binary Search Tree root nodes to start
* processing. This constructor creates two trees from two given sorted
* array inputs. root1 tree from arr1 and root2 tree from arr2.
*
* Once we are done with creating the tree, we are traversing the tree in
* inorder format, to verify whether nodes are inserted properly or not. An
* inorder traversal should give us the nodes in a sorted order.
*/
public MergeTwoBST_to_BalancedBST() {
// passing 0 as the startIndex and arr1.length-1 as the endIndex.
root1 = getBSTFromSortedArray(arr1, 0, arr1.length - 1);
System.out.println("\nPrinting the first binary search tree");
inorder(root1); // traverse the tree in inorder format to verify whether
// nodes are inserted correctly or not.
// passing 0 as the startIndex and arr2.length-1 as the endIndex.
root2 = getBSTFromSortedArray(arr2, 0, arr2.length - 1);
System.out.println("\nPrinting the second binary search tree");
inorder(root2); // same here - checking whether the nodes are inserted
// properly or not.
}
/**
* Method to traverse the tree in inorder format. Where it traverses the
* left child first, then root and then right child.
*
* #param node
*/
public void inorder(BSTNode node) {
if (null != node) {
inorder(node.getLeft());
System.out.print(node.getData() + " ");
inorder(node.getRight());
}
}
/**
* Method to traverse the tree in preorder format. Where it traverses the
* root node first, then left child and then right child.
*
* #param node
*/
public void preorder(BSTNode node) {
if (null != node) {
System.out.print(node.getData() + " ");
preorder(node.getLeft());
preorder(node.getRight());
}
}
/**
* Creating a new Binary Search Tree object from a sorted array and
* returning the root of the newly created node for further processing.
*
* #param arr
* #param startIndex
* #param endIndex
* #return
*/
public BSTNode getBSTFromSortedArray(int[] arr, int startIndex, int endIndex) {
if (startIndex > endIndex) {
return null;
}
int middleIndex = startIndex + (endIndex - startIndex) / 2;
BSTNode node = new BSTNode(arr[middleIndex]);
node.setLeft(getBSTFromSortedArray(arr, startIndex, middleIndex - 1));
node.setRight(getBSTFromSortedArray(arr, middleIndex + 1, endIndex));
return node;
}
/**
* This method involves two operation. First - it adds the nodes from root1
* tree to root2 tree, and hence we get a merged root2 tree.Second - it
* balances the merged root2 tree with the help of a vector object which can
* contain objects only of BSTNode type.
*/
public void mergeTwoBinarySearchTree() {
// First operation - merging the trees. root1 with root2 merging should
// give us a new root2 tree.
addUtil(root1);
System.out.println("\nAfter the root1 tree nodes are added to root2");
System.out.println("Inorder Traversal of root2 nodes");
inorder(root2); // inorder traversal of the root2 tree should display
// the nodes in a sorted order.
System.out.println("\nPreorder traversal of root2 nodes");
preorder(root2);
// Second operation - this will take care of balancing the merged binary
// search trees.
balancedTheMergedBST();
}
/**
* Here we are doing two operations. First operation involves, adding nodes
* from root2 tree to the vector object. Second operation involves, creating
* the Balanced binary search tree from the vector objects.
*/
public void balancedTheMergedBST() {
// First operation : adding nodes to the vector object
addNodesToVector(root2, vNodes);
int vSize = vNodes.size();
// Second operation : getting a balanced binary search tree
BSTNode node = getBalancedBSTFromVector(vNodes, 0, vSize - 1);
System.out
.println("\n********************************************************");
System.out.println("After balancing the merged trees");
System.out.println("\nInorder Traversal of nodes");
inorder(node); // traversing the tree in inoder process should give us
// the output in sorted order ascending
System.out.println("\nPreorder traversal of root2 nodes");
preorder(node);
}
/**
* This method will provide us a Balanced Binary Search Tree. Elements of
* the root2 tree has been added to the vector object. It is parsed
* recursively to create a balanced tree.
*
* #param vNodes
* #param startIndex
* #param endIndex
* #return
*/
public BSTNode getBalancedBSTFromVector(Vector<BSTNode> vNodes,
int startIndex, int endIndex) {
if (startIndex > endIndex) {
return null;
}
int middleIndex = startIndex + (endIndex - startIndex) / 2;
BSTNode node = vNodes.get(middleIndex);
node.setLeft(getBalancedBSTFromVector(vNodes, startIndex,
middleIndex - 1));
node.setRight(getBalancedBSTFromVector(vNodes, middleIndex + 1,
endIndex));
return node;
}
/**
* This method traverse the tree in inorder process and adds each node from
* root2 to the vector object vNodes object only accepts objects of BSTNode
* type.
*
* #param node
* #param vNodes
*/
public void addNodesToVector(BSTNode node, Vector<BSTNode> vNodes) {
if (null != node) {
addNodesToVector(node.getLeft(), vNodes);
// here we are adding the node in the vector object.
vNodes.add(node);
addNodesToVector(node.getRight(), vNodes);
}
}
/**
* This method traverse the root1 tree in inorder process and add the nodes
* in the root2 tree based on their value
*
* #param node
*/
public void addUtil(BSTNode node) {
if (null != node) {
addUtil(node.getLeft());
mergeToSecondTree(root2, node.getData());
addUtil(node.getRight());
}
}
/**
* This method adds the nodes found from root1 tree as part it's inorder
* traversal and add it to the second tree.
*
* This method follows simple Binary Search Tree inserstion logic to insert
* a node considering the tree already exists.
*
* #param node
* #param data
* #return
*/
public BSTNode mergeToSecondTree(BSTNode node, int data) {
if (null == node) {
node = new BSTNode(data);
} else {
if (data < node.getData()) {
node.setLeft(mergeToSecondTree(node.getLeft(), data));
} else if (data > node.getData()) {
node.setRight(mergeToSecondTree(node.getRight(), data));
}
}
return node;
}
/**
*
* #param args
*/
public static void main(String[] args) {
MergeTwoBST_to_BalancedBST mergeTwoBST = new MergeTwoBST_to_BalancedBST();
mergeTwoBST.mergeTwoBinarySearchTree();
}
}
BSTNode.java:
public class BSTNode {
BSTNode left, right;
int data;
/* Default constructor */
public BSTNode() {
left = null;
right = null;
data = 0;
}
/* Constructor */
public BSTNode(int data) {
left = null;
right = null;
this.data = data;
}
public BSTNode getLeft() {
return left;
}
public void setLeft(BSTNode left) {
this.left = left;
}
public BSTNode getRight() {
return right;
}
public void setRight(BSTNode right) {
this.right = right;
}
public int getData() {
return data;
}
public void setData(int data) {
this.data = data;
}
}