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 
 

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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

 

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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. 
 

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.