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:

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

// 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; 
} 

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: