Merge two BSTs with limited extra space | GeeksforGeeks
Given two Binary Search Trees(BST), print the elements of both BSTs in sorted form. The expected time complexity is O(m+n) where m is the number of nodes in first tree and n is the number of nodes in second tree. Maximum allowed auxiliary space is O(height of the first tree + height of the second tree).
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.
Time Complexity: O(m+n)
Auxiliary Space: O(height of the first tree + height of the second tree)
Read full article from Merge two BSTs with limited extra space | GeeksforGeeks
Given two Binary Search Trees(BST), print the elements of both BSTs in sorted form. The expected time complexity is O(m+n) where m is the number of nodes in first tree and n is the number of nodes in second tree. Maximum allowed auxiliary space is O(height of the first tree + height of the second tree).
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.
Time Complexity: O(m+n)
Auxiliary Space: O(height of the first tree + height of the second tree)
void merge(struct node *root1, struct node *root2){ // s1 is stack to hold nodes of first BST struct snode *s1 = NULL; // Current node of first BST struct node *current1 = root1; // s2 is stack to hold nodes of second BST struct snode *s2 = NULL; // Current node of second BST struct node *current2 = root2; // If first BST is empty, then output is inorder // traversal of second BST if (root1 == NULL) { inorder(root2); return; } // If second BST is empty, then output is inorder // traversal of first BST if (root2 == NULL) { inorder(root1); return ; } // Run the loop while there are nodes not yet printed. // The nodes may be in stack(explored, but not printed) // or may be not yet explored while (current1 != NULL || !isEmpty(s1) || current2 != NULL || !isEmpty(s2)) { // Following steps follow iterative Inorder Traversal if (current1 != NULL || current2 != NULL ) { // Reach the leftmost node of both BSTs and push ancestors of // leftmost nodes to stack s1 and s2 respectively if (current1 != NULL) { push(&s1, current1); current1 = current1->left; } if (current2 != NULL) { push(&s2, current2); current2 = current2->left; } } else { // If we reach a NULL node and either of the stacks is empty, // then one tree is exhausted, ptint the other tree if (isEmpty(s1)) { while (!isEmpty(s2)) { current2 = pop (&s2); current2->left = NULL; inorder(current2); } return ; } if (isEmpty(s2)) { while (!isEmpty(s1)) { current1 = pop (&s1); current1->left = NULL; inorder(current1); } return ; } // Pop an element from both stacks and compare the // popped elements current1 = pop(&s1); current2 = pop(&s2); // If element of first tree is smaller, then print it // and push the right subtree. If the element is larger, // then we push it back to the corresponding stack. if (current1->data < current2->data) { printf("%d ", current1->data); current1 = current1->right; push(&s2, current2); current2 = NULL; } else { printf("%d ", current2->data); current2 = current2->right; push(&s1, current1); current1 = NULL; } } }}