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# Number of Binary Trees for given Preorder Sequence length

• Difficulty Level : Medium
• Last Updated : 23 Jun, 2022

Count the number of Binary Tree possible for a given Preorder Sequence length n.
Examples:

```Input : n = 1
Output : 1

Input : n = 2
Output : 2

Input : n = 3
Output : 5```

Background :

In Preorder traversal, we process the root node first, then traverse the left child node and then right child node.
For example preorder traversal of below tree is 1 2 4 5 3 6 7 Finding number of trees with given Preorder:

Number of Binary Tree possible if such a traversal length (let’s say n) is given.
Let’s take an Example : Given Preorder Sequence –> 2 4 6 8 10 (length 5).

• Assume there is only 1 node (that is 2 in this case), So only 1 Binary tree is Possible
• Now, assume there are 2 nodes (namely 2 and 4), So only 2 Binary Tree are Possible:
• Now, when there are 3 nodes (namely 2, 4 and 6), So Possible Binary tree are 5
• Consider 4 nodes (that are 2, 4, 6 and 8), So Possible Binary Tree are 14.
Let’s say BT(1) denotes number of Binary tree for 1 node. (We assume BT(0)=1)
BT(4) = BT(0) * BT(3) + BT(1) * BT(2) + BT(2) * BT(1) + BT(3) * BT(0)
BT(4) = 1 * 5 + 1 * 2 + 2 * 1 + 5 * 1 = 14
• Similarly, considering all the 5 nodes (2, 4, 6, 8 and 10). Possible number of Binary Tree are:
BT(5) = BT(0) * BT(4) + BT(1) * BT(3) + BT(2) * BT(2) + BT(3) * BT(1) + BT(4) * BT(0)
BT(5) = 1 * 14 + 1 * 5 + 2 * 2 + 5 * 1 + 14 * 1 = 42 Hence, Total binary Tree for Pre-order sequence of length 5 is 42.
We use Dynamic programming to calculate the possible number of Binary Tree. We take one node at a time and calculate the possible Trees using previously calculated Trees.

## C++

 `// C++ Program to count possible binary trees``// using dynamic programming``#include ``using` `namespace` `std;` `int` `countTrees(``int` `n)``{``    ``// Array to store number of Binary tree``    ``// for every count of nodes``    ``int` `BT[n + 1];``    ``memset``(BT, 0, ``sizeof``(BT));` `    ``BT = BT = 1;` `    ``// Start finding from 2 nodes, since``    ``// already know for 1 node.``    ``for` `(``int` `i = 2; i <= n; ++i)``        ``for` `(``int` `j = 0; j < i; j++)``            ``BT[i] += BT[j] * BT[i - j - 1];` `    ``return` `BT[n];``}` `// Driver code``int` `main()``{``    ``int` `n = 5;``    ``cout << ``"Total Possible Binary Trees are : "``        ``<< countTrees(n) << endl;``    ``return` `0;``}`

## Java

 `// Java Program to count``// possible binary trees``// using dynamic programming``import` `java.io.*;` `class` `GFG``{``static` `int` `countTrees(``int` `n)``{``    ``// Array to store number``    ``// of Binary tree for``    ``// every count of nodes``    ``int` `BT[] = ``new` `int``[n + ``1``];``    ``for``(``int` `i = ``0``; i <= n; i++)``    ``BT[i] = ``0``;``    ``BT[``0``] = BT[``1``] = ``1``;` `    ``// Start finding from 2``    ``// nodes, since already``    ``// know for 1 node.``    ``for` `(``int` `i = ``2``; i <= n; ++i)``        ``for` `(``int` `j = ``0``; j < i; j++)``            ``BT[i] += BT[j] *``                     ``BT[i - j - ``1``];` `    ``return` `BT[n];``}` `// Driver code``public` `static` `void` `main (String[] args)``{``int` `n = ``5``;``System.out.println(``"Total Possible "` `+``                ``"Binary Trees are : "` `+``                       ``countTrees(n));``}``}` `// This code is contributed by anuj_67.`

## Python3

 `# Python3 Program to count possible binary``# trees using dynamic programming` `def` `countTrees(n) :` `    ``# Array to store number of Binary``    ``# tree for every count of nodes``    ``BT ``=` `[``0``] ``*` `(n ``+` `1``)` `    ``BT[``0``] ``=` `BT[``1``] ``=` `1` `    ``# Start finding from 2 nodes, since``    ``# already know for 1 node.``    ``for` `i ``in` `range``(``2``, n ``+` `1``):``        ``for` `j ``in` `range``(i):``            ``BT[i] ``+``=` `BT[j] ``*` `BT[i ``-` `j ``-` `1``]` `    ``return` `BT[n]` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:` `    ``n ``=` `5``    ``print``(``"Total Possible Binary Trees are : "``,``                                 ``countTrees(n))``                                 ` `# This code is contributed by``# Shubham Singh(SHUBHAMSINGH10)`

## C#

 `// C# Program to count``// possible binary trees``// using dynamic programming``using` `System;` `class` `GFG``{``static` `int` `countTrees(``int` `n)``{``    ``// Array to store number``    ``// of Binary tree for``    ``// every count of nodes``    ``int` `[]BT = ``new` `int``[n + 1];``    ``for``(``int` `i = 0; i <= n; i++)``        ``BT[i] = 0;``        ``BT = BT = 1;` `    ``// Start finding from 2``    ``// nodes, since already``    ``// know for 1 node.``    ``for` `(``int` `i = 2; i <= n; ++i)``        ``for` `(``int` `j = 0; j < i; j++)``            ``BT[i] += BT[j] *``                     ``BT[i - j - 1];` `    ``return` `BT[n];``}` `// Driver code``static` `public` `void` `Main (String []args)``{``    ``int` `n = 5;``    ``Console.WriteLine(``"Total Possible "` `+``                      ``"Binary Trees are : "` `+``                             ``countTrees(n));``}``}` `// This code is contributed``// by Arnab Kundu`

## PHP

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## Javascript

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Output:

`Total Possible Binary Trees are : 42`

Time Complexity: O(n2)

Auxiliary Space : O(n)

Alternative :
This can also be done using Catalan number Cn = (2n)!/(n+1)!*n!
For n = 0, 1, 2, 3, … values of Catalan numbers are 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, …. So are numbers of Binary Search Trees.

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