A binary tree is full if all of its vertices have either zero or two children. Let B_n denote the number of full binary trees with n vertices. Implement a function to find B_n.

To find the number of full binary trees with n vertices , we can use dynamic programming . We can use a two - dimensional array to store the number of full binary trees with each number of vertices in the left sub tree and each number of vertices in the right sub tree . \n \n We can initialize the first row and column of the array to 1 , since a tree with no vertices has one full binary tree . Then , for each row i and column j , we can calculate the number of full binary trees in the it h row and j th column as follows : \n \n * If i = 0 , then the number of full binary trees in the it h row and j th column is the number of full binary trees with j - 1 vertices in the left sub tree and j - 1 vertices in the right sub tree . We can calculate this by recurs ively calling the function with n - 1 and j - 1 . \n * If i = 1 , then the number of full binary trees in the it h row and j th column is the number of full binary trees with n - 1 vertices in the left sub tree and j - 1 vertices in the right sub tree . We can calculate this by recurs ively calling the function with n - 1 and j - 1 . \n * If i 0 , then the number of full binary trees in the it h row and j th column is the sum of the number of full binary trees in the it h row and j - 1 column and the number of full binary trees in the it h row and j + 1 column . We can calculate this by recurs ively calling the function with n - 1 and j - 1 , and n - 1 and j + 1 . \n \n We can then calculate the total number of full binary trees by sum ming the values in the last row and last column of the array . \n \n Here is the implementation of the function in Python : \n ``` \n def count _ full _ trees ( n ): \n # Initialize the first row and column to 1 \n count = [ 1 ] * ( n + 1 ) \n count [ 0 ] = count [ 1 ] = 1 \n \n # Calcul ate the number of full binary trees in each row and column \n for i in range ( 2 , n + 1 ): \n for j in range ( 1 , n + 1 ): \n if i == 0 : \n count [ i ][ j ] = count [ n - 1 ][ j - 1 ] \n elif i == 1 : \n count [ i ][ j ] = count [ n - 1 ][ j - 1 ] \n else : \n count [ i ][ j ] = count [ i - 1 ][ j - 1 ] + count [ i - 1 ][ j + 1 ] \n \n # Calcul ate the total number of full binary trees \n total = sum ( count [ i ][ n ]) \n \n return total \n ``` \n For example , if we call the function with n = 3 , it will return the following : \n ``` \n count _ full _ trees ( 3 ) \n # Output : 5 \n ``` \n This is because there are 5 full binary trees with 3 vertices : \n ``` \n 1 \n / \ \n 2 3 \n / \ / \ \n 4 5 6 7 \n ``` \n We can also use a recursive approach to find the number of full binary trees with n vertices . We can define a function ` count _ full _ trees _ rec ursive ` that takes a positive integer n and returns the number of full binary trees with n vertices . We can then call this function with n = 1 , 2 , 3 , ... , n to calculate the number of full binary trees with each number of vertices . \n \n Here is the implementation of the recursive function in Python : \n ``` \n def count _ full _ trees _ rec ursive ( n ): \n # Base case : there is one full binary tree with 0 vertices \n if n == 0 : \n return 1 \n \n # Rec ursive case : there are n - 1 full binary trees with n - 1 vertices in the left sub tree and n - 1 vertices in the right sub tree \n return count _ full _ trees _ rec ursive ( n - 1 ) + count _ full _ trees _ rec ursive ( n - 1 ) \n ``` \n We can then call this function with n = 1 , 2 , 3 , ... , n to calculate the number of full binary trees with each number of vertices . \n \n For example , if we call the function with n = 3 , it will return the following : \n ``` \n count _ full _ trees _ rec ursive ( 3 ) \n # Output : 5 \n ``` \n This is because there are 5 full binary trees with 3 vertices : \n ``` \n 1 \n / \ \n 2 3 \n / \ / \ \n