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

Last Updated : 11 Sep, 2023

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"```

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 ` `// subsequence 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[0]); ` ` `  `    ``seqExist(root, seq, n)? cout << ``"Yes"` `: ` `                            ``cout << ``"No"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find if given array  ` `// exists as a subsequence 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 subsequence 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[0]]  ` `    ``# 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 `

## C#

 `// C# program to find if given array  ` `// exists as a subsequence in BST  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `// A binary Tree node  ` `public` `class` `Node  ` `{  ` `    ``public` `int` `data;  ` `    ``public` `Node left, right;  ` `};  ` ` `  `// structure of int class  ` `public` `class` `INT  ` `{  ` `    ``public` `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` `bool` `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))  ` `        ``Console.WriteLine(``"Yes"``);  ` `    ``else` `        ``Console.WriteLine(``"No"``);  ` `}  ` `}  ` ` `  `/* This code contributed by PrinciRaj1992 */`

## Javascript

 ``

Output

`Yes`

Time complexity: O(n)
Auxiliary Space: O(n) for call stack since using recursion, where n is total no of nodes in BST

Approach 2 :- Optimal solution in O(N) time complexity and O(h) space complexity

The existsSubsequence function takes a BST root and a vector seq representing the sorted sub-sequence to be searched. It first checks if the seq is empty, in which case it returns true. Otherwise, it calls the findSequence function with the root, seq, and index i set to 0. The findSequence function recursively searches for the elements in the seq starting from index i in the BST. If the current node has the same value as the element at index i, it increments i and recursively calls the findSequence function on the left and right subtrees with the updated value of i. If i reaches the end of seq, it means that the sub-sequence exists in the BST, and the function returns true. If the current node does not have the same value as the element at index i, the function returns false.

## C++

 `#include ` `#include ` ` `  `using` `namespace` `std; ` ` `  `struct` `TreeNode { ` `    ``int` `val; ` `    ``TreeNode* left; ` `    ``TreeNode* right; ` `    ``TreeNode(``int` `x) : val(x), left(NULL), right(NULL) {} ` `}; ` ` `  `bool` `findSequence(TreeNode* root, vector<``int``>& seq, ``int` `i) { ` `    ``if` `(root == NULL) { ` `        ``return` `false``; ` `    ``} ` `    ``if` `(root->val == seq[i]) { ` `        ``i++; ` `        ``if` `(i == seq.size()) { ` `            ``return` `true``; ` `        ``} ` `        ``return` `findSequence(root->left, seq, i) || findSequence(root->right, seq, i); ` `    ``} ` `    ``return` `false``; ` `} ` ` `  `bool` `existsSubsequence(TreeNode* root, vector<``int``>& seq) { ` `    ``if` `(seq.empty()) { ` `        ``return` `true``; ` `    ``} ` `    ``return` `findSequence(root, seq, 0); ` `} ` ` `  `int` `main() { ` `    ``// create a BST ` `    ``TreeNode* root = ``new` `TreeNode(5); ` `    ``root->left = ``new` `TreeNode(3); ` `    ``root->right = ``new` `TreeNode(7); ` `    ``root->left->left = ``new` `TreeNode(2); ` `    ``root->left->right = ``new` `TreeNode(4); ` `    ``root->right->left = ``new` `TreeNode(6); ` `    ``root->right->right = ``new` `TreeNode(8); ` ` `  `    ``// check if a given sorted sub-sequence exists in the BST ` `    ``vector<``int``> seq1 = {2, 3, 4}; ` `    `  `    ``if` `(existsSubsequence(root, seq1)) { ` `        ``cout << ``"The sequence {2, 3, 4} exists in the BST"` `<< endl; ` `    ``} ``else` `{ ` `        ``cout << ``"The sequence {2, 3, 4} does not exist in the BST"` `<< endl; ` `    ``} ` `     `  ` `  `    ``return` `0; ` `} `

## Java

 `import` `java.util.ArrayList; ` ` `  `class` `TreeNode { ` `    ``int` `val; ` `    ``TreeNode left; ` `    ``TreeNode right; ` `     `  `    ``public` `TreeNode(``int` `x) { ` `        ``val = x; ` `        ``left = ``null``; ` `        ``right = ``null``; ` `    ``} ` `} ` ` `  `public` `class` `Main { ` `    ``public` `static` `boolean` `findSequence(TreeNode root, ArrayList seq, ``int` `i) { ` `        ``if` `(root == ``null``) { ` `            ``return` `false``; ` `        ``} ` `        ``if` `(root.val == seq.get(i)) { ` `            ``i++; ` `            ``if` `(i == seq.size()) { ` `                ``return` `true``; ` `            ``} ` `            ``return` `findSequence(root.left, seq, i) || findSequence(root.right, seq, i); ` `        ``} ` `        ``return` `false``; ` `    ``} ` `     `  `    ``public` `static` `boolean` `existsSubsequence(TreeNode root, ArrayList seq) { ` `        ``if` `(seq.isEmpty()) { ` `            ``return` `true``; ` `        ``} ` `        ``return` `findSequence(root, seq, ``0``); ` `    ``} ` `     `  `    ``public` `static` `void` `main(String[] args) { ` `        ``// create a BST ` `        ``TreeNode root = ``new` `TreeNode(``5``); ` `        ``root.left = ``new` `TreeNode(``3``); ` `        ``root.right = ``new` `TreeNode(``7``); ` `        ``root.left.left = ``new` `TreeNode(``2``); ` `        ``root.left.right = ``new` `TreeNode(``4``); ` `        ``root.right.left = ``new` `TreeNode(``6``); ` `        ``root.right.right = ``new` `TreeNode(``8``); ` `     `  `        ``// check if a given sorted sub-sequence exists in the BST ` `        ``ArrayList seq1 = ``new` `ArrayList<>(); ` `        ``seq1.add(``2``); ` `        ``seq1.add(``3``); ` `        ``seq1.add(``4``); ` `         `  `        ``if` `(existsSubsequence(root, seq1)) { ` `            ``System.out.println(``"The sequence {2, 3, 4} exists in the BST"``); ` `        ``} ``else` `{ ` `            ``System.out.println(``"The sequence {2, 3, 4} does not exist in the BST"``); ` `        ``} ` `    ``} ` `} `

## Python3

 `# javascript code implementation  ` `class` `Node: ` `     `  `    ``def` `__init__(``self``, val): ` `        ``self``.val ``=` `val ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` `     `  ` `  `def` `findSequence(root, seq, i): ` `    ``if` `(root ``=``=` `None``): ` `        ``return` `False` `     `  `    ``if` `(root.val ``=``=` `seq[i]): ` `        ``i ``=` `i ``+` `1` `        ``if` `(i ``=``=` `len``(seq)): ` `            ``return` `True` `        ``return` `findSequence(root.left, seq, i) ``or` `findSequence(root.right, seq, i) ` `    ``return` `False` ` `  ` `  `def` `existsSubsequence(root, seq): ` `    ``if` `(``len``(seq) ``=``=` `0``): ` `        ``return` `True` `    ``return` `findSequence(root, seq, ``0``) ` ` `  ` `  `root ``=` `Node(``5``) ` `root.left ``=` `Node(``3``) ` `root.right ``=` `Node(``7``) ` `root.left.left ``=` `Node(``2``) ` `root.left.right ``=` `Node(``4``) ` `root.right.left ``=` `Node(``6``) ` `root.right.right ``=` `Node(``8``) ` ` `  `# check if a given sorted sub-sequence exists in the BST ` `seq1 ``=` `[``2``, ``3``, ``4``] ` ` `  `if` `(existsSubsequence(root, seq1)): ` `    ``print``(``"The sequence {2, 3, 4} exists in the BST"``) ` `else``: ` `    ``print``(``"The sequence {2, 3, 4} does not exist in the BST"``) ` ` `  `# The code is contributed by Nidhi goel.  `

## C#

 `using` `System; ` `using` `System.Collections.Generic; ` ` `  `public` `class` `TreeNode { ` `    ``public` `int` `val; ` `    ``public` `TreeNode left; ` `    ``public` `TreeNode right; ` `    ``public` `TreeNode(``int` `x) { val = x; } ` `} ` ` `  `public` `class` `Program { ` `    ``public` `static` `bool` `FindSequence(TreeNode root, ` `                                    ``List<``int``> seq, ``int` `i) ` `    ``{ ` `        ``if` `(root == ``null``) { ` `            ``return` `false``; ` `        ``} ` `        ``if` `(root.val == seq[i]) { ` `            ``i++; ` `            ``if` `(i == seq.Count) { ` `                ``return` `true``; ` `            ``} ` `            ``return` `FindSequence(root.left, seq, i) ` `                ``|| FindSequence(root.right, seq, i); ` `        ``} ` `        ``return` `false``; ` `    ``} ` ` `  `    ``public` `static` `bool` `ExistsSubsequence(TreeNode root, ` `                                         ``List<``int``> seq) ` `    ``{ ` `        ``if` `(seq.Count == 0) { ` `            ``return` `true``; ` `        ``} ` `        ``return` `FindSequence(root, seq, 0); ` `    ``} ` ` `  `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``// create a BST ` `        ``TreeNode root = ``new` `TreeNode(5); ` `        ``root.left = ``new` `TreeNode(3); ` `        ``root.right = ``new` `TreeNode(7); ` `        ``root.left.left = ``new` `TreeNode(2); ` `        ``root.left.right = ``new` `TreeNode(4); ` `        ``root.right.left = ``new` `TreeNode(6); ` `        ``root.right.right = ``new` `TreeNode(8); ` ` `  `        ``// check if a given sorted sub-sequence exists in ` `        ``// the BST ` `        ``List<``int``> seq1 = ``new` `List<``int``>{ 2, 3, 4 }; ` ` `  `        ``if` `(ExistsSubsequence(root, seq1)) { ` `            ``Console.WriteLine( ` `                ``"The sequence {2, 3, 4} exists in the BST"``); ` `        ``} ` `        ``else` `{ ` `            ``Console.WriteLine( ` `                ``"The sequence {2, 3, 4} does not exist in the BST"``); ` `        ``} ` `    ``} ` `} ` `// This code is contributed by user_dtewbxkn77n`

## Javascript

 `// javascript code implementation  ` `class Node { ` `     `  `    ``constructor(val){ ` `        ``this``.val = val; ` `        ``this``.left = ``null``; ` `        ``this``.right = ``null``; ` `    ``} ` `} ` ` `  `function` `findSequence(root, seq, i) { ` `    ``if` `(root == ``null``) { ` `        ``return` `false``; ` `    ``} ` `    ``if` `(root.val == seq[i]) { ` `        ``i++; ` `        ``if` `(i == seq.length) { ` `            ``return` `true``; ` `        ``} ` `        ``return` `findSequence(root.left, seq, i) || findSequence(root.right, seq, i); ` `    ``} ` `    ``return` `false``; ` `} ` ` `  `function` `existsSubsequence(root, seq) { ` `    ``if` `(seq.length == 0) { ` `        ``return` `true``; ` `    ``} ` `    ``return` `findSequence(root, seq, 0); ` `} ` ` `  ` `  `let root = ``new` `Node(5); ` `root.left = ``new` `Node(3); ` `root.right = ``new` `Node(7); ` `root.left.left = ``new` `Node(2); ` `root.left.right = ``new` `Node(4); ` `root.right.left = ``new` `Node(6); ` `root.right.right = ``new` `Node(8); ` ` `  `// check if a given sorted sub-sequence exists in the BST ` `let seq1 = [2, 3, 4]; ` ` `  `if` `(existsSubsequence(root, seq1)) { ` `    ``console.log(``"The sequence {2, 3, 4} exists in the BST"``); ` `} ``else` `{ ` `    ``console.log(``"The sequence {2, 3, 4} does not exist in the BST"``); ` `} ` ` `  `// The code is contributed by Arushi Jindal.`

Output

`The sequence {2, 3, 4} does not exist in the BST`

Time complexity –The time complexity of the given program depends on the height of the BST and the length of the input sequence. In the worst case, where the input sequence is a valid subsequence of the BST and has a length of k, and the height of the BST is h, the time complexity of the program will be O(k*h).

The reason for this is that the program performs a depth-first search on the BST to find the nodes in the input sequence, and the search may need to visit up to k nodes in the BST. Since the search is a recursive function that traverses the tree from the root to the leaves, the worst-case time complexity is O(k*h).

Space complexity – The space complexity of the program depends on the height of the BST and the length of the input sequence. In the worst case, where the input sequence is a valid subsequence of the BST and has a length of k, and the height of the BST is h, the space complexity of the program will be O(h).

The reason for this is that the program uses a recursive function to traverse the BST, and the maximum number of function calls that can be on the call stack at any given time is equal to the height of the BST. In the worst case, where the input sequence is a valid subsequence of the BST and has a length of k, and the height of the BST is h, the space complexity is O(h).

Previous Article
Next Article
Article Tags :
Practice Tags :