Given a binary search tree and a sorted sub-sequence. the task is to check if the given sorted sub-sequence exist in binary search tree or not.

Examples:

// For above binary search tree Input : seq[] = {4, 6, 8, 14} Output: "Yes" Input : seq[] = {1, 2, 3, 8} Output: "No"

A **simple solution** is to store **inorder traversal** in an auxiliary array and then by matching elements of sorted sub-sequence one by one with inorder traversal of tree , we can if sub-sequence exist in BST or not. Time complexity for this approach will be O(n) but it requires extra space O(n) for storing traversal in an array.

An **efficient solution** is to match elements of sub-sequence while we are traversing BST in **inorder** fashion. We take **index** as a iterator for given sorted sub-sequence and start inorder traversal of given bst, if **current node** matches with **seq[index]** then move **index** in forward direction by incrementing 1 and after complete traversal of BST if **index==n** that means all elements of given sub-sequence have been matched and exist as a sorted sub-sequence in given BST.

// C++ program to find if given array exists as a // subsequece in BST #include<bits/stdc++.h> using namespace std; // A binary Tree node struct Node { int data; struct Node *left, *right; }; // A utility function to create a new BST node // with key as given num struct Node* newNode(int num) { struct Node* temp = new Node; temp->data = num; temp->left = temp->right = NULL; return temp; } // A utility function to insert a given key to BST struct Node* insert(struct Node* root, int key) { if (root == NULL) return newNode(key); if (root->data > key) root->left = insert(root->left, key); else root->right = insert(root->right, key); return root; } // function to check if given sorted sub-sequence exist in BST // index --> iterator for given sorted sub-sequence // seq[] --> given sorted sub-sequence void seqExistUtil(struct Node *ptr, int seq[], int &index) { if (ptr == NULL) return; // We traverse left subtree first in Inorder seqExistUtil(ptr->left, seq, index); // If current node matches with se[index] then move // forward in sub-sequence if (ptr->data == seq[index]) index++; // We traverse left subtree in the end in Inorder seqExistUtil(ptr->right, seq, index); } // A wrapper over seqExistUtil. It returns true // if seq[0..n-1] exists in tree. bool seqExist(struct Node *root, int seq[], int n) { // Initialize index in seq[] int index = 0; // Do an inorder traversal and find if all // elements of seq[] were present seqExistUtil(root, seq, index); // index would become n if all elements of // seq[] were present return (index == n); } // driver program to run the case int main() { struct Node* root = NULL; root = insert(root, 8); root = insert(root, 10); root = insert(root, 3); root = insert(root, 6); root = insert(root, 1); root = insert(root, 4); root = insert(root, 7); root = insert(root, 14); root = insert(root, 13); int seq[] = {4, 6, 8, 14}; int n = sizeof(seq)/sizeof(seq[0]); seqExist(root, seq, n)? cout << "Yes" : cout << "No"; return 0; }

Output:

Yes

Time complexity : O(n)

This article is contributed by **Shashank Mishra ( Gullu )**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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