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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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)

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C#

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// 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

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PHP

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<?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), "\n"
  
// This code is contributed by ajit.
?>

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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.

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