Diameter of an N-ary tree

The diameter of an N-ary tree is the longest path present between any two nodes of the tree. These two nodes must be two leaf nodes. The following examples have the longest path[diameter] shaded.

Examples:
diameternary

Example 2:
diametern2



Prerequisite : Diameter of a binary tree.

The path can either start from one of the node and goes up to one of the LCAs of these nodes and again come down to the deepest node of some other subtree or can exist as a diameter of one of the child of the current node.
The solution will exist in any one of these:
I] Diameter of one of the children of the current node
II] Sum of Height of highest two subtree + 1

// C++ program to find the height of an N-ary
// tree
#include <bits/stdc++.h>
using namespace std;

// Structure of a node of an n-ary tree
struct Node
{
    char key;
    vector<Node *> child;
};

// Utility function to create a new tree node
Node *newNode(int key)
{
    Node *temp = new Node;
    temp->key = key;
    return temp;
}

// Utility function that will return the depth
// of the tree
int depthOfTree(struct Node *ptr)
{
    // Base case
    if (!ptr)
        return 0;

    int maxdepth = 0;

    // Check for all children and find
    // the maximum depth
    for (vector<Node*>::iterator it = ptr->child.begin();
                           it != ptr->child.end(); it++)

        maxdepth = max(maxdepth , depthOfTree(*it));

    return maxdepth + 1;
}

// Function to calculate the diameter
// of the tree
int diameter(struct Node *ptr)
{
    // Base case
    if (!ptr)
        return 0;

    // Find top two highest children
    int max1 = 0, max2 = 0;
    for (vector<Node*>::iterator it = ptr->child.begin();
                          it != ptr->child.end(); it++)
    {
        int h = depthOfTree(*it);
        if (h > max1)
           max2 = max1, max1 = h;
        else if (h > max2)
           max2 = h;
    }

    // Iterate over each child for diameter
    int maxChildDia = 0;
    for (vector<Node*>::iterator it = ptr->child.begin();
                           it != ptr->child.end(); it++)
        maxChildDia = max(maxChildDia, diameter(*it));

    return max(maxChildDia, max1 + max2 + 1);
}

// Driver program
int main()
{
    /*   Let us create below tree
    *           A
    *         / /  \  \
    *       B  F   D  E
    *      / \     |  /|\
    *     K  J    G  C H I
    *      /\            \
    *    N   M            L
    */

    Node *root = newNode('A');
    (root->child).push_back(newNode('B'));
    (root->child).push_back(newNode('F'));
    (root->child).push_back(newNode('D'));
    (root->child).push_back(newNode('E'));
    (root->child[0]->child).push_back(newNode('K'));
    (root->child[0]->child).push_back(newNode('J'));
    (root->child[2]->child).push_back(newNode('G'));
    (root->child[3]->child).push_back(newNode('C'));
    (root->child[3]->child).push_back(newNode('H'));
    (root->child[3]->child).push_back(newNode('I'));
    (root->child[0]->child[0]->child).push_back(newNode('N'));
    (root->child[0]->child[0]->child).push_back(newNode('M'));
    (root->child[3]->child[2]->child).push_back(newNode('L'));

    cout << diameter(root) << endl;

    return 0;
}

Output:



7

Optimizations to above solution :
We can make a hash table to store heights of all nodes. If we precompute these heights, we don’t need to call depthOfTree() for every node.

A different optimized solution :
Longest path in an undirected tree

This article is contributed by Shubham Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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