# 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:

- https://www.geeksforgeeks.org/diameter-n-ary-tree/
- https://www.geeksforgeeks.org/diameter-n-ary-tree-using-bfs/

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; ` `} ` |

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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

## Recommended Posts:

- Diameter of an N-ary tree
- Diameter of n-ary tree using BFS
- Diameter of a tree using DFS
- Diameter of a Binary Tree
- Diameter of a Binary Tree in O(n) [A new method]
- Possible edges of a tree for given diameter, height and vertices
- Make a tree with n vertices , d diameter and at most vertex degree k
- Find number of edges that can be broken in a tree such that Bitwise OR of resulting two trees are equal
- Count the Number of Binary Search Trees present in a Binary Tree
- Generic Trees(N-array Trees)
- Total number of possible Binary Search Trees and Binary Trees with n keys
- Maximum sub-tree sum in a Binary Tree such that the sub-tree is also a BST
- Complexity of different operations in Binary tree, Binary Search Tree and AVL tree
- Given level order traversal of a Binary Tree, check if the Tree is a Min-Heap
- Convert an arbitrary Binary Tree to a tree that holds Children Sum Property

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