human
Can you provide a code to count of subtrees from an N and the other steps required given the following:
Count of subtrees from an N | Function to implement DFS traversal.
Mark node v as visited.
Traverse Adj_List of node v.
If current node is not visited.
DFS call for current node.
Count the total red and blue nodes of children of its subtree.
Count the no . of red and blue nodes in the subtree.
If subtree contains all red node & no blue node.
If subtree contains all blue node & no red node.
Function to count the number of nodes with red color.
Function to count the number of nodes with blue color.
Function to create a Tree with given vertices.
Traverse the edge [ ] array.
Create adjacency list.
Function to count the number of subtree with the given condition.
For creating adjacency list.
To store the count of subtree with only blue and red color.
visited array for DFS Traversal.
Count the number of red node in the tree.
Count the number of blue node in the tree.
Function Call to build tree.
DFS Traversal.
Print the final count.
Driver Code"
How can the above be solved in Python?
assistant
#Here is code to count of subtrees from an N and the other steps required.
 def Solution_dfs ( v , color , red , blue , sub_red , sub_blue , vis , adj , ans ) :
	vis [ v ] = 1 ;
	for i in range ( len ( adj [ v ] ) ) :
		if ( vis [ adj [ v ] [ i ] ] == 0 ) :
			ans = Solution_dfs ( adj [ v ] [ i ] , color , red , blue , sub_red , sub_blue , vis , adj , ans ) ;
			sub_red [ v ] += sub_red [ adj [ v ] [ i ] ] ;
			sub_blue [ v ] += sub_blue [ adj [ v ] [ i ] ] ;
	if ( color [ v ] == 1 ) :
		sub_red [ v ] += 1 ;
	if ( color [ v ] == 2 ) :
		sub_blue [ v ] += 1 ;
	if ( sub_red [ v ] == red and sub_blue [ v ] == 0 ) :
		( ans ) += 1 ;
	if ( sub_red [ v ] == 0 and sub_blue [ v ] == blue ) :
		( ans ) += 1 ;
	return ans
def countRed ( color , n ) :
	red = 0 ;
	for i in range ( n ) :
		if ( color [ i ] == 1 ) :
			red += 1 ;
	return red ;
def countBlue ( color , n ) :
	blue = 0 ;
	for i in range ( n ) :
		if ( color [ i ] == 2 ) :
			blue += 1
	return blue ;
def buildTree ( edge , m , n ) :
	u , v , i = 0 , 0 , 0
	for i in range ( n - 1 ) :
		u = edge [ i ] [ 0 ] - 1 ;
		v = edge [ i ] [ 1 ] - 1 ;
		if u not in m :
			m [ u ] = [ ]
		if v not in m :
			m [ v ] = [ ]
		m [ u ] . append ( v )
		m [ v ] . append ( u ) ;
def countSubtree ( color , n , edge ) :
	adj = dict ( )
	ans = 0 ;
	sub_red = [ 0 for i in range ( n + 3 ) ]
	sub_blue = [ 0 for i in range ( n + 3 ) ]
	vis = [ 0 for i in range ( n + 3 ) ]
	red = countRed ( color , n ) ;
	blue = countBlue ( color , n ) ;
	buildTree ( edge , adj , n ) ;
	ans = Solution_dfs ( 0 , color , red , blue , sub_red , sub_blue , vis , adj , ans ) ;
	print ( ans , end = ' ' )
if __name__ == ' _ _ main _ _ ' :
	N = 5 ;
	color = [ 1 , 0 , 0 , 0 , 2 ]
	edge = [ [ 1 , 2 ] , [ 2 , 3 ] , [ 3 , 4 ] , [ 4 , 5 ] ] ;
	countSubtree ( color , N , edge ) ;