Minimum valued node having maximum depth in an N-ary Tree

Given a tree of N nodes, the task is to find the node having maximum depth starting from the root node, taking the root node at zero depth. If there are more than 1 maximum depth node, then find the one having the smallest value. 

Examples: 

Input:     
             1
           /   \
          2     3
         /  \
        4    5

Output: 4
Explanation:
For this tree: 
Height of Node 1 - 0, 
Height of Node 2 - 1, 
Height of Node 3 - 1, 
Height of Node 4 - 2, 
Height of Node 5 - 2. 
Hence, the nodes whose height is 
maximum are 4 and 5, out of which 
4 is minimum valued.

Input:     
             1
           /   
          2   
         /
        3  

Output: 3
Explanation:
For this tree: 
Height of Node 1 - 0, 
Height of Node 2 - 1, 
Height of Node 3 - 2
Hence, the node whose height 
is maximum is 3.


Approach: 

  • The idea is to use Depth First Search(DFS) on the tree and for every node, check the height of every node as we move down the tree.
  • Check if it is the maximum so far or not and if it has a height equal to the maximum value, then is it the minimum valued node or not.
  • If yes then update the maximum height so far and the node value accordingly.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of for
// the above problem
 
#include <bits/stdc++.h>
using namespace std;
 
#define MAX 100000
 
vector<int> graph[MAX + 1];
 
// To store the height of each node
int maxHeight, minNode;
 
// Function to perform dfs
void dfs(int node, int parent,
         int h)
{
    // Store the height of node
    int height = h;
 
    if (height > maxHeight) {
        maxHeight = height;
        minNode = node;
    }
    else if (height == maxHeight
             && minNode > node)
        minNode = node;
 
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node, h + 1);
    }
}
 
// Driver code
int main()
{
    // Number of nodes
    int N = 5;
 
    // Edges of the tree
    graph[1].push_back(2);
    graph[1].push_back(3);
    graph[2].push_back(4);
    graph[2].push_back(5);
 
    maxHeight = 0;
    minNode = 1;
 
    dfs(1, 1, 0);
 
    cout << minNode << "\n";
 
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of for
// the above problem
import java.util.*;
 
class GFG{
 
static final int MAX = 100000;
@SuppressWarnings("unchecked")
static Vector<Integer>[] graph = new Vector[MAX + 1];
 
// To store the height of each node
static int maxHeight, minNode;
 
// Function to perform dfs
static void dfs(int node, int parent, int h)
{
     
    // Store the height of node
    int height = h;
    if (height > maxHeight)
    {
        maxHeight = height;
        minNode = node;
    }
    else if (height == maxHeight &&
             minNode > node)
        minNode = node;
 
    for(int to : graph[node])
    {
        if (to == parent)
            continue;
        dfs(to, node, h + 1);
    }
}
 
// Driver code
public static void main(String[] args)
{
    // Number of nodes
    int N = 5;
    for(int i = 0; i < graph.length; i++)
        graph[i] = new Vector<Integer>();
     
    // Edges of the tree
    graph[1].add(2);
    graph[1].add(3);
    graph[2].add(4);
    graph[2].add(5);
    maxHeight = 0;
    minNode = 1;
    dfs(1, 1, 0);
    System.out.print(minNode + "\n");
}
}
 
// This code is contributed by sapnasingh4991

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of for
# the above problem
MAX = 100000
  
graph = [[] for i in range(MAX + 1)]
  
# To store the height of each node
maxHeight = 0
minNode = 0
  
# Function to perform dfs
def dfs(node, parent, h):
     
    global minNode, maxHeight
     
    # Store the height of node
    height = h
  
    if (height > maxHeight):
        maxHeight = height
        minNode = node
     
    elif (height == maxHeight and
          minNode > node):
        minNode = node
     
    for to in graph[node]:
        if to == parent:
            continue
         
        dfs(to, node, h + 1)
         
# Driver code
if __name__=="__main__":
     
    # Number of nodes
    N = 5
  
    # Edges of the tree
    graph[1].append(2)
    graph[1].append(3)
    graph[2].append(4)
    graph[2].append(5)
  
    maxHeight = 0
    minNode = 1
  
    dfs(1, 1, 0)
     
    print(minNode)
 
# This code is contributed by rutvik_56

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of for
// the above problem
using System;
using System.Collections.Generic;
  
public class GFG{
  
static readonly int MAX = 100000;
static List<int>[] graph = new List<int>[MAX + 1];
  
// To store the height of each node
static int maxHeight, minNode;
  
// Function to perform dfs
static void dfs(int node, int parent, int h)
{
      
    // Store the height of node
    int height = h;
    if (height > maxHeight)
    {
        maxHeight = height;
        minNode = node;
    }
    else if (height == maxHeight &&
             minNode > node)
        minNode = node;
  
    foreach(int to in graph[node])
    {
        if (to == parent)
            continue;
        dfs(to, node, h + 1);
    }
}
  
// Driver code
public static void Main(String[] args)
{
    for(int i = 0; i < graph.Length; i++)
        graph[i] = new List<int>();
          
    // Edges of the tree
    graph[1].Add(2);
    graph[1].Add(3);
    graph[2].Add(4);
    graph[2].Add(5);
    maxHeight = 0;
    minNode = 1;
    dfs(1, 1, 0);
    Console.Write(minNode + "\n");
}
}
  
// This code is contributed by shikhasingrajput

chevron_right


Output: 

4





 

 

competitive-programming-img




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.