DP on Trees | Set-3 ( Diameter of N-ary Tree )

Given an N-ary tree T of N nodes, the task is to calculate the longest path between any two nodes(also known as the diameter of the tree).

Example 1:

Example 2:

Different approaches to solve these problems have already been discussed:

In this post, we will be discussing an approach which uses Dynamic Programming on Trees.

Prerequisites:

There are two possibilities for the diameter to exist:

  • Case 1: Suppose the diameter starts from a node and ends at some node in its subtree. Let’s say that there exist a node x such that the longest path starts from node x and goes into its subtree and ends at some node in the subtree itself. Let’s define this path length by dp1[x].
  • Case 2: Suppose the diameter or the longest path starts in subtree of a node x, passes through it and ends in it’s subtree. Let’s define this path by dp2[x].

If for all nodes x, we take a maximum of dp1[x], dp2[x], then we will get the diameter of the tree.

For the case-1, to find dp1[node], we need to find the maximum of all dp1[x], where x is the children of node. And dp1[node] will be equal to 1 + max(dp1[children1], dp1[children2], ..).

For the case-2, to find dp2[node], we need to find the two maximum of all dp1[x], where x is the children of node. And dp2[node] will be equal to 1 + max 2 of(dp1[children1], dp1[children2], ..).

We can easily run a DFS and find the maximum of both dp1[node] and dp2[node] for every to get the diameter of the tree.

Below is the implementation of the above approach:

C++

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// C++ program to find diameter of a tree
// using DFS.
#include <bits/stdc++.h>
using namespace std;
  
int diameter = -1;
  
// Function to find the diameter of the tree
// using Dynamic Programming
int dfs(int node, int parent, int dp1[], int dp2[], list<int>* adj)
{
  
    // Store the first maximum and secondmax
    int firstmax = -1;
    int secondmax = -1;
  
    // Traverse for all children of node
    for (auto i = adj[node].begin(); i != adj[node].end(); ++i) {
        if (*i == parent)
            continue;
  
        // Call DFS function again
        dfs(*i, node, dp1, dp2, adj);
  
        // Find first max
        if (firstmax == -1) {
            firstmax = dp1[*i];
        }
        else if (dp1[*i] >= firstmax) // Secondmaximum
        {
            secondmax = firstmax;
            firstmax = dp1[*i];
        }
        else if (dp1[*i] > secondmax) // Find secondmaximum
        {
            secondmax = dp1[*i];
        }
    }
  
    // Base case for every node
    dp1[node] = 1;
    if (firstmax != -1) // Add
        dp1[node] += firstmax;
  
    // Find dp[2]
    if (secondmax != -1)
        dp2[node] = 1 + firstmax + secondmax;
  
    // Return maximum of both
    return max(dp1[node], dp2[node]);
}
  
// Driver Code
int main()
{
    int n = 5;
  
    /* Constructed tree is 
         1 
        / \ 
        2 3 
       / \ 
       4  5 */
    list<int>* adj = new list<int>[n + 1];
  
    /*create undirected edges */
    adj[1].push_back(2);
    adj[2].push_back(1);
    adj[1].push_back(3);
    adj[3].push_back(1);
    adj[2].push_back(4);
    adj[4].push_back(2);
    adj[2].push_back(5);
    adj[5].push_back(2);
  
    int dp1[n + 1], dp2[n + 1];
    memset(dp1, 0, sizeof dp1);
    memset(dp2, 0, sizeof dp2);
  
    // Find diameter by calling function
    cout << "Diameter of the given tree is "
         << dfs(1, 1, dp1, dp2, adj) << endl;
  
    return 0;
}

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Python3





# Python3 program to find diameter

# of a tree using DFS. 


# Function to find the diameter of the

# tree using Dynamic Programming

def dfs(node, parent, dp1, dp2, adj): 


	# Store the first maximum and secondmax

	firstmax, secondmax = -1, -1


	# Traverse for all children of node

	for i in adj[node]:

		if i == parent:

			continue


		# Call DFS function again

		dfs(i, node, dp1, dp2, adj) 


		# Find first max

		if firstmax == -1:

			firstmax = dp1[i] 


		elif dp1[i] >= firstmax: # Secondmaximum

			secondmax = firstmax

			firstmax = dp1[i] 


		elif dp1[i] > secondmax: # Find secondmaximum

			secondmax = dp1[i] 


	# Base case for every node

	dp1[node] = 1

	if firstmax != -1: # Add

		dp1[node] += firstmax 


	# Find dp[2]

	if secondmax != -1:

		dp2[node] = 1 + firstmax + secondmax 


	# Return maximum of both

	return max(dp1[node], dp2[node]) 


# Driver Code

if __name__ == “__main__”:


	n, diameter = 5, -1


	adj = [[] for i in range(n + 1)]


	# create undirected edges

	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) 


	# Find diameter by calling function

	print(“Diameter of the given tree is”,

		         dfs(1, 1, dp1, dp2, adj))


# This code is contributed by Rituraj Jain 





Output:


Diameter of the given tree is 4






          
          
          
          

            

    

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