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"
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<bits/stdc++.h>using namespace std; // A binary Tree nodestruct Node{ int data; struct Node *left, *right;}; // A utility function to create a new BST node// with key as given numstruct 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 BSTstruct 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-sequencevoid 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 caseint 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 subsequece in BST import java.util.*; class GFG{ // A binary Tree node static class Node { int data; Node left, right; }; //structure of int classstatic 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 BSTclass 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 Codeif __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 subsequece 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 */ |
Output:
Yes
Time complexity: O(n)
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