# Check if given sorted sub-sequence exists in binary search tree

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[] = {4, 6, 8, 12, 13}
Output: "No"
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to find if given array exists as a ` `// subsequece in BST ` `#include ` `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); ` ` `  `    ``seqExist(root, seq, n)? cout << ``"Yes"` `: ` `                            ``cout << ``"No"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find if given array  ` `// exists as a subsequece in BST  ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `     `  `// A binary Tree node  ` `static` `class` `Node  ` `{  ` `    ``int` `data;  ` `    ``Node left, right;  ` `};  ` ` `  `//structure of int class ` `static` `class` `INT ` `{ ` `    ``int` `a; ` `} ` ` `  `// A utility function to create a new BST node  ` `// with key as given num  ` `static` `Node newNode(``int` `num)  ` `{  ` `    ``Node temp = ``new` `Node();  ` `    ``temp.data = num;  ` `    ``temp.left = temp.right = ``null``;  ` `    ``return` `temp;  ` `}  ` ` `  `// A utility function to insert a given key to BST  ` `static` `Node insert( 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  ` `static` `void` `seqExistUtil( 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.a])  ` `        ``index.a++;  ` ` `  `    ``// 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.  ` `static` `boolean` `seqExist( Node root, ``int` `seq[], ``int` `n)  ` `{  ` `    ``// Initialize index in seq[]  ` `    ``INT index = ``new` `INT(); ` `     `  `    ``index.a = ``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.a == n);  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String args[])  ` `{  ` `    ``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 = seq.length;  ` ` `  `    ``if``(seqExist(root, seq, n)) ` `        ``System.out.println(``"Yes"``); ` `    ``else` `        ``System.out.println(``"No"``); ` `}  ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## Python3

 `# Python3 program to find if given array ` `# exists as a subsequece in BST ` `class` `Node:  ` ` `  `    ``# Constructor to create a new node  ` `    ``def` `__init__(``self``, data):  ` `        ``self``.data ``=` `data  ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` `         `  `# A utility function to insert a  ` `# given key to BST  ` `def` `insert(root, key): ` `    ``if` `root ``=``=` `None``:  ` `        ``return` `Node(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  ` `def` `seqExistUtil(ptr, seq, index): ` `    ``if` `ptr ``=``=` `None``:  ` `        ``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[``0``]]:  ` `        ``index[``0``] ``+``=` `1` ` `  `    ``# 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.  ` `def` `seqExist(root, seq, n): ` `     `  `    ``# Initialize index in seq[]  ` `    ``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  ` `    ``if` `index[``0``] ``=``=` `n: ` `        ``return` `True` `    ``else``: ` `        ``return` `False` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``root ``=` `None` `    ``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``) ` ` `  `    ``seq ``=` `[``4``, ``6``, ``8``, ``14``]  ` `    ``n ``=` `len``(seq) ` `    ``if` `seqExist(root, seq, n): ` `        ``print``(``"Yes"``) ` `    ``else``: ` `        ``print``(``"No"``)  ` ` `  `# This code is contributed by PranchalK `

Output:

```Yes
```

Time complexity: O(n)