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
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
NOTE* Since we have already calculated for 1, 2 and 3 nodes. We don’t need to evaluate them again for successive nodes.
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.
Total Possible Binary Trees are : 42
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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