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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rakesh

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