You are given two balanced binary search trees e.g., AVL or Red Black Tree. Write a function that merges the two given balanced BSTs into a balanced binary search tree. Let there be m elements in first tree and n elements in the other tree. Your merge function should take O(m+n) time.

In the following solutions, it is assumed that sizes of trees are also given as input. If the size is not given, then we can get the size by traversing the tree (See this).

**Method 1 (Insert elements of first tree to second) **

Take all elements of first BST one by one, and insert them into the second BST. Inserting an element to a self balancing BST takes Logn time (See this) where n is size of the BST. So time complexity of this method is Log(n) + Log(n+1) … Log(m+n-1). The value of this expression will be between mLogn and mLog(m+n-1). As an optimization, we can pick the smaller tree as first tree.

**Method 2 (Merge Inorder Traversals) **

1) Do inorder traversal of first tree and store the traversal in one temp array arr1[]. This step takes O(m) time.

2) Do inorder traversal of second tree and store the traversal in another temp array arr2[]. This step takes O(n) time.

3) The arrays created in step 1 and 2 are sorted arrays. Merge the two sorted arrays into one array of size m + n. This step takes O(m+n) time.

4) Construct a balanced tree from the merged array using the technique discussed in this post. This step takes O(m+n) time.

Time complexity of this method is O(m+n) which is better than method 1. This method takes O(m+n) time even if the input BSTs are not balanced.

Following is C++ implementation of this method.

#include <stdio.h> #include <stdlib.h> /* A binary tree node has data, pointer to left child and a pointer to right child */ struct node { int data; struct node* left; struct node* right; }; // A utility unction to merge two sorted arrays into one int *merge(int arr1[], int arr2[], int m, int n); // A helper function that stores inorder traversal of a tree in inorder array void storeInorder(struct node* node, int inorder[], int *index_ptr); /* A function that constructs Balanced Binary Search Tree from a sorted array See http://www.geeksforgeeks.org/archives/17138 */ struct node* sortedArrayToBST(int arr[], int start, int end); /* This function merges two balanced BSTs with roots as root1 and root2. m and n are the sizes of the trees respectively */ struct node* mergeTrees(struct node *root1, struct node *root2, int m, int n) { // Store inorder traversal of first tree in an array arr1[] int *arr1 = new int[m]; int i = 0; storeInorder(root1, arr1, &i); // Store inorder traversal of second tree in another array arr2[] int *arr2 = new int[n]; int j = 0; storeInorder(root2, arr2, &j); // Merge the two sorted array into one int *mergedArr = merge(arr1, arr2, m, n); // Construct a tree from the merged array and return root of the tree return sortedArrayToBST (mergedArr, 0, m+n-1); } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct node* newNode(int data) { struct node* node = (struct node*) malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return(node); } // A utility function to print inorder traversal of a given binary tree void printInorder(struct node* node) { if (node == NULL) return; /* first recur on left child */ printInorder(node->left); printf("%d ", node->data); /* now recur on right child */ printInorder(node->right); } // A utility unction to merge two sorted arrays into one int *merge(int arr1[], int arr2[], int m, int n) { // mergedArr[] is going to contain result int *mergedArr = new int[m + n]; int i = 0, j = 0, k = 0; // Traverse through both arrays while (i < m && j < n) { // Pick the smaler element and put it in mergedArr if (arr1[i] < arr2[j]) { mergedArr[k] = arr1[i]; i++; } else { mergedArr[k] = arr2[j]; j++; } k++; } // If there are more elements in first array while (i < m) { mergedArr[k] = arr1[i]; i++; k++; } // If there are more elements in second array while (j < n) { mergedArr[k] = arr2[j]; j++; k++; } return mergedArr; } // A helper function that stores inorder traversal of a tree rooted with node void storeInorder(struct node* node, int inorder[], int *index_ptr) { if (node == NULL) return; /* first recur on left child */ storeInorder(node->left, inorder, index_ptr); inorder[*index_ptr] = node->data; (*index_ptr)++; // increase index for next entry /* now recur on right child */ storeInorder(node->right, inorder, index_ptr); } /* A function that constructs Balanced Binary Search Tree from a sorted array See http://www.geeksforgeeks.org/archives/17138 */ struct node* sortedArrayToBST(int arr[], int start, int end) { /* Base Case */ if (start > end) return NULL; /* Get the middle element and make it root */ int mid = (start + end)/2; struct node *root = newNode(arr[mid]); /* Recursively construct the left subtree and make it left child of root */ root->left = sortedArrayToBST(arr, start, mid-1); /* Recursively construct the right subtree and make it right child of root */ root->right = sortedArrayToBST(arr, mid+1, end); return root; } /* Driver program to test above functions*/ int main() { /* Create following tree as first balanced BST 100 / \ 50 300 / \ 20 70 */ struct node *root1 = newNode(100); root1->left = newNode(50); root1->right = newNode(300); root1->left->left = newNode(20); root1->left->right = newNode(70); /* Create following tree as second balanced BST 80 / \ 40 120 */ struct node *root2 = newNode(80); root2->left = newNode(40); root2->right = newNode(120); struct node *mergedTree = mergeTrees(root1, root2, 5, 3); printf ("Following is Inorder traversal of the merged tree \n"); printInorder(mergedTree); getchar(); return 0; }

Output:

Following is Inorder traversal of the merged tree 20 40 50 70 80 100 120 300

**Method 3 (In-Place Merge using DLL)**

We can use a Doubly Linked List to merge trees in place. Following are the steps.

1) Convert the given two Binary Search Trees into doubly linked list in place (Refer this post for this step).

2) Merge the two sorted Linked Lists (Refer this post for this step).

3) Build a Balanced Binary Search Tree from the merged list created in step 2. (Refer this post for this step)

Time complexity of this method is also O(m+n) and this method does conversion in place.

Thanks to Dheeraj and Ronzii for suggesting this method.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.