human
Give me a code to dP on Trees and the other steps required given the following:
DP on Trees | Set | Function to find the diameter of the tree using Dynamic Programming.
Store the first maximum and secondmax.
Traverse for all children of node.
Call DFS function again.
Find first max.
Secondmaximum.
elif dp1 [ i ] > secondmax : Find secondmaximum.
Base case for every node.
if firstmax != - 1 : Add.
Find dp [ 2 ].
Return maximum of both.
Driver Code.
Constructed tree is 1 / \ 2 3 / \ 4 5.
create undirected edges.
Find diameter by calling function"
How can the above be solved in Python?
assistant
#Here is code to dP on Trees and the other steps required.
 def dfs ( node , parent , dp1 , dp2 , adj ) :
	firstmax , secondmax = - 1 , - 1
	for i in adj [ node ] :
		if i == parent :
			continue
		dfs ( i , node , dp1 , dp2 , adj )
		if firstmax == - 1 :
			firstmax = dp1 [ i ]
	elif dp1 [ i ] >= firstmax :
		secondmax = firstmax
		firstmax = dp1 [ i ]
		secondmax = dp1 [ i ]
	dp1 [ node ] = 1
		dp1 [ node ] += firstmax
	if secondmax != - 1 :
		dp2 [ node ] = 1 + firstmax + secondmax
	return max ( dp1 [ node ] , dp2 [ node ] )
if __name__ == " _ _ main _ _ " :
	n , diameter = 5 , - 1
	adj = [ [ ] for i in range ( n + 1 ) ]
	adj [ 1 ] . append ( 2 )
	adj [ 2 ] . append ( 1 )
	adj [ 1 ] . append ( 3 )
	adj [ 3 ] . append ( 1 )
	adj [ 2 ] . append ( 4 )
	adj [ 4 ] . append ( 2 )
	adj [ 2 ] . append ( 5 )
	adj [ 5 ] . append ( 2 )
	dp1 = [ 0 ] * ( n + 1 )
	dp2 = [ 0 ] * ( n + 1 )
	print ( " Diameter ▁ of ▁ the ▁ given ▁ tree ▁ is " , dfs ( 1 , 1 , dp1 , dp2 , adj ) )