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Number of Binary Trees for given Preorder Sequence length

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

  • 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++

// 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 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[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
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

<?php
// PHP Program to count possible binary 
// trees using dynamic programming 
  
function countTrees($n
    // Array to store number of Binary 
    // tree for every count of nodes 
    $BT[$n + 1] = array(); 
    $BT = array_fill(0, $n + 1, NULL);
    $BT[0] = $BT[1] = 1; 
  
    // Start finding from 2 nodes, since 
    // already know for 1 node. 
    for ($i = 2; $i <= $n; ++$i
        for ($j = 0; $j < $i; $j++) 
            $BT[$i] += $BT[$j] * 
                       $BT[$i - $j - 1]; 
  
    return $BT[$n]; 
  
// Driver code 
$n = 5; 
echo "Total Possible Binary Trees are : ",
                    countTrees($n), " "
  
// This code is contributed by ajit.
?>

Output:

Total Possible Binary Trees 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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