# Check if two given key sequences construct same BSTs

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 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
```

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

1) Compare sizes of two arrays. If not same, return false.
2) Compare first values of two arrays. If not same, return false.
3) 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.
4) Recursively check the first list of first array with first list of second and same for second list.

## 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 aL1, ` `                            ``ArrayList 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 aLLeft1 = ``new` `ArrayList(); ` `        ``ArrayList aLRight1 = ``new` `ArrayList(); ` `        ``ArrayList aLLeft2 = ``new` `ArrayList(); ` `        ``ArrayList aLRight2 = ``new` `ArrayList(); ` `        ``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 aL1 = ` `                         ``new` `ArrayList(Arrays. ` `                           ``asList(``3``, ``5``, ``4``, ``6``, ``1``, ``0``, ``2``)); ` `        ``ArrayList aL2 =  ` `                         ``new` `ArrayList(Arrays. ` `                          ``asList(``3``, ``1``, ``5``, ``2``, ``4``, ``6``, ``0``)); ` ` `  `        ``System.out.println(sameBSTs(aL1, aL2)); ` `    ``} ` `} `

## 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 != aL2) ` `            ``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) ` `                ``aLLeft1.Add(aL1[i]); ` `            ``else` `                ``aLRight1.Add(aL1[i]); ` ` `  `            ``if` `(aL2[i] < aL2) ` `                ``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 */`

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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Improved By : princiraj1992