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++ 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 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 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# 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 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.
?>
<script>
// Javascript Program to count
// possible binary trees
// using dynamic programming
function countTrees(n)
{
// Array to store number
// of Binary tree for
// every count of nodes
let BT = new Array(n + 1);
for(let 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 (let i = 2; i <= n; ++i)
for (let j = 0; j < i; j++)
BT[i] += BT[j] * BT[i - j - 1];
return BT[n];
}
let n = 5;
document.write("Total Possible " + "Binary Trees are : " + countTrees(n));
</script>
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.