Merge Two Balanced Binary Search Trees

Source

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;

}