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 |
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Output:
Diameter of the given tree is 4
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- Diameter of an N-ary tree
- Diameter of a tree using DFS
- Diameter of n-ary tree using BFS
- 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
- Diameter of a Binary Indexed Tree with N nodes
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- 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
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Improved By : rituraj_jain
