# Number of Binary Trees for given Preorder Sequence length

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

```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

NOTE* Since we have already calculated for 1, 2 and 3 nodes. We don’t need to evaluate them again for successive nodes.

• 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++ Program to count possible binary trees`
`// using dynamic programming`
`#include <bits/stdc++.h>`
`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[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`
`int` `main()`
`{`
`    ``int` `n = 5;`
`    ``cout << ``"Total Possible Binary Tree are : "`
`        ``<< countTrees(n) << endl;`
`    ``return` `0;`
`}`

Output:

```Total Possible Binary Tree are : 42
```

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.

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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