Given two arrays which represent two sequences of keys that are used to create BSTs. Imagine we make a Binary Search Tree (BST) from each array. We need to tell whether two BSTs will be identical or not without actually constructing the tree.
Examples:
Let the input arrays be a[] and b[]
Example 1:
a[] = {2, 4, 1, 3} will construct following tree.
2
/ \
1 4
/
3
b[] = {2, 4, 3, 1} will also construct the same tree.
2
/ \
1 4
/
3
So the output is "True"
Example 2:
a[] = {8, 3, 6, 1, 4, 7, 10, 14, 13}
b[] = {8, 10, 14, 3, 6, 4, 1, 7, 13}
They both construct the same following BST, so output is "True"
8
/ \
3 10
/ \ \
1 6 14
/ \ /
4 7 13
- Compare sizes of two arrays. If not same, return false.
- Compare first values of two arrays. If not same, return false.
- Create two lists from each given array such that the first list contains values smaller than first item of the corresponding array. And second list contains greater values.
- Recursively check the first list of first array with the first list of the second and same for second list.
Implementation:
C++
// C++ program to check if given two arrays represent// same BST.#include <bits/stdc++.h>using namespace std;bool sameBSTs(vector<int> aL1, vector<int> aL2){ // Base cases if (aL1.size() != aL2.size()) return false; if (aL1.size() == 0) return true; if (aL1[0] != aL2[0]) return false; // Construct two lists from each input array. The first // list contains values smaller than first value, i.e., // left subtree. And second list contains right subtree. vector<int> aLLeft1 ; vector<int> aLRight1 ; vector<int> aLLeft2 ; vector<int> aLRight2 ; for (int i = 1; i < aL1.size(); i++) { if (aL1[i] < aL1[0]) aLLeft1.push_back(aL1[i]); else aLRight1.push_back(aL1[i]); if (aL2[i] < aL2[0]) aLLeft2.push_back(aL2[i]); else aLRight2.push_back(aL2[i]); } // Recursively compare left and right // subtrees. return sameBSTs(aLLeft1, aLLeft2) && sameBSTs(aLRight1, aLRight2);}// Driver codeint main(){ vector<int> aL1; vector<int> aL2; aL1.push_back(3); aL1.push_back(5); aL1.push_back(4); aL1.push_back(6); aL1.push_back(1); aL1.push_back(0); aL1.push_back(2); aL2.push_back(3); aL2.push_back(1); aL2.push_back(5); aL2.push_back(2); aL2.push_back(4); aL2.push_back(6); aL2.push_back(0); cout << ((sameBSTs(aL1, aL2))?"true":"false")<<"\n"; return 0;}// This code is contributed by Arnab Kundu |
Java
// Java program to check if given two arrays represent// same BST.import java.util.ArrayList;import java.util.Arrays;public class SameBST { static boolean sameBSTs(ArrayList<Integer> aL1, ArrayList<Integer> aL2) { // Base cases if (aL1.size() != aL2.size()) return false; if (aL1.size() == 0) return true; if (aL1.get(0) != aL2.get(0)) return false; // Construct two lists from each input array. The first // list contains values smaller than first value, i.e., // left subtree. And second list contains right subtree. ArrayList<Integer> aLLeft1 = new ArrayList<Integer>(); ArrayList<Integer> aLRight1 = new ArrayList<Integer>(); ArrayList<Integer> aLLeft2 = new ArrayList<Integer>(); ArrayList<Integer> aLRight2 = new ArrayList<Integer>(); for (int i = 1; i < aL1.size(); i++) { if (aL1.get(i) < aL1.get(0)) aLLeft1.add(aL1.get(i)); else aLRight1.add(aL1.get(i)); if (aL2.get(i) < aL2.get(0)) aLLeft2.add(aL2.get(i)); else aLRight2.add(aL2.get(i)); } // Recursively compare left and right // subtrees. return sameBSTs(aLLeft1, aLLeft2) && sameBSTs(aLRight1, aLRight2); } // Driver code public static void main(String[] args) { ArrayList<Integer> aL1 = new ArrayList<Integer>(Arrays. asList(3, 5, 4, 6, 1, 0, 2)); ArrayList<Integer> aL2 = new ArrayList<Integer>(Arrays. asList(3, 1, 5, 2, 4, 6, 0)); System.out.println(sameBSTs(aL1, aL2)); }} |
Python3
# Python3 program to check if given two arrays represent# same BST.def sameBSTs(aL1, aL2): # Base cases if (len(aL1) != len(aL2)): return False if (len(aL1) == 0): return True if (aL1[0] != aL2[0]): return False # Construct two lists from each input array. The first # list contains values smaller than first value, i.e., # left subtree. And second list contains right subtree. aLLeft1 = [] aLRight1 = [] aLLeft2 = [] aLRight2 = [] for i in range(1, len(aL1)): if (aL1[i] < aL1[0]): aLLeft1.append(aL1[i]) else: aLRight1.append(aL1[i]) if (aL2[i] < aL2[0]): aLLeft2.append(aL2[i]) else: aLRight2.append(aL2[i]) # Recursively compare left and right # subtrees. return sameBSTs(aLLeft1, aLLeft2) and sameBSTs(aLRight1, aLRight2)# Driver codeaL1 = []aL2 = []aL1.append(3)aL1.append(5)aL1.append(4)aL1.append(6)aL1.append(1)aL1.append(0)aL1.append(2)aL2.append(3)aL2.append(1)aL2.append(5)aL2.append(2)aL2.append(4)aL2.append(6)aL2.append(0)if (sameBSTs(aL1, aL2)): print("true")else: print("false")# This code is contributed by shubhamsingh10 |
C#
// C# program to check if given // two arrays represent same BST.using System;using System.Linq;using System.Collections.Generic;public class SameBST { static bool sameBSTs(List<int> aL1, List<int> aL2) { // Base cases if (aL1.Count != aL2.Count) return false; if (aL1.Count == 0) return true; if (aL1[0] != aL2[0]) return false; // Construct two lists from each // input array. The first list contains // values smaller than first value, i.e., // left subtree. And second list contains right subtree. List<int> aLLeft1 = new List<int>(); List<int> aLRight1 = new List<int>(); List<int> aLLeft2 = new List<int>(); List<int> aLRight2 = new List<int>(); for (int i = 1; i < aL1.Count; i++) { if (aL1[i] < aL1[0]) aLLeft1.Add(aL1[i]); else aLRight1.Add(aL1[i]); if (aL2[i] < aL2[0]) aLLeft2.Add(aL2[i]); else aLRight2.Add(aL2[i]); } // Recursively compare left and right // subtrees. return sameBSTs(aLLeft1, aLLeft2) && sameBSTs(aLRight1, aLRight2); } // Driver code public static void Main() { int []arr1 = {3, 5, 4, 6, 1, 0, 2}; List<int> aL1 = arr1.ToList(); int []arr2 = {3, 1, 5, 2, 4, 6, 0}; List<int> aL2 = arr2.ToList(); Console.WriteLine(sameBSTs(aL1, aL2)); }}/* This code contributed by PrinciRaj1992 */ |
Javascript
<script>// javascript program to check if given two arrays represent// same BST.function sameBSTs(aL1, aL2){ // Base cases if (aL1.length != aL2.length) return false; if (aL1.length == 0) return true; if (aL1[0] !== aL2[0]) return false; // Construct two lists from each input array. The first // list contains values smaller than first value, i.e., // left subtree. And second list contains right subtree. var aLLeft1 = []; var aLRight1 =[] ; var aLLeft2 =[]; var aLRight2 =[]; for (var i = 1; i < aL1.length; i++) { if (aL1[i] < aL1[0]) aLLeft1.push(aL1[i]); else aLRight1.push(aL1[i]); if (aL2[i] < aL2[0]) aLLeft2.push(aL2[i]); else aLRight2.push(aL2[i]); } // Recursively compare left and right // subtrees. if(sameBSTs(aLLeft1, aLLeft2) && sameBSTs(aLRight1, aLRight2)) { return true; } return false;}// Driver codevar aL1 =[3, 5, 4, 6, 1, 0, 2] ;var aL2 =[3, 1, 5, 2, 4, 6, 0];document.write( ((sameBSTs(aL1, aL2))?"true":"false")+"<br>");</script> |
Output
true
Time Complexity: O(n * n)
Please refer below post for O(n) solution
Check for Identical BSTs without building the trees
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