Given a Tree with N nodes values from 1 to N and N – 1 edges. The task is to find the maximum matching in the given tree.
A matching in a tree is a collection of edges such that no pair of edges share a common node. Matching with the most edges is known as a maximum matching.
Examples:
Input: Below is the given graph:
Output: 3
Explanation:
Set of Edges in the above graph for maximum matching:
(4, 5), (1, 2), (7, 8)Input: Below is the given graph:
Output: 3
Explanation:
Set of Edges in the above graph for maximum matching:
(4, 5), (2, 3), (1, 7)
Approach: This problem can be solved using Greedy Approach and the idea is to use post-order traversal in the tree and start with leaf edges and go up the order. Below are the steps:
- Perform DFS Traversal on the given tree with rooted node 1 and make parent as 0 and pass current nodes as a parent of the node in recursive DFS traversal.
- While performing traversal, for each node U and its parent node P if these nodes are unvisited then mark these nodes as visited and increment the maximum matching count by 1.
- Print the count of a maximum matching in the above step after DFS Traversal.
The Greedy algorithm is to repeatedly take any leaf-edge.
TreeMatch(F:forest) M <- [] while F nonempty do { select any leaf-edge e M <- M + [e] F <- F - both ends of e }
Why the greedy algorithm works correctly?
Let’s, assume E is a leaf edge and consider any maximum matching N. Suppose N does not contain E. Then if we add E to N, only one vertex now has two edges incident with it. So we can delete one of the edges of N and attain a maximum matching containing E.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
#define N 10000 // Adjacency list to store edges vector< int > adj[N];
int used[N];
int max_matching;
// Add an edge between U and V in tree void AddEdge( int u, int v)
{ // Edge from u to v
adj[u].push_back(v);
// Edge from V to U
adj[v].push_back(u);
} // Function that finds the maximum // matching of the DFS void Matching_dfs( int u, int p)
{ for ( int i = 0;
i < adj[u].size(); i++) {
// Go further as we are not
// allowed to go towards
// its parent
if (adj[u][i] != p) {
Matching_dfs(adj[u][i], u);
}
}
// If U and its parent P is
// not taken then we must
// take &mark them as taken
if (!used[u] and !used[p] and p != 0) {
// Increment size of edge set
max_matching++;
used[u] = used[p] = 1;
}
} // Function to find the maximum // matching in a graph void maxMatching()
{ // Taking 1 as a root of the tree
Matching_dfs(1, 0);
// Print maximum Matching
cout << max_matching << "\n" ;
} // Driver Code int main()
{ int n = 5;
// Joining edge between
// two nodes in tree
AddEdge(1, 2);
AddEdge(1, 3);
AddEdge(3, 4);
AddEdge(3, 5);
// Function Call
maxMatching();
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
static final int N = 10000 ;
// Adjacency list to store edges @SuppressWarnings ( "unchecked" )
static Vector<Integer>[] adj = new Vector[N];
static int used[] = new int [N];
static int max_matching;
// Add an edge between U and V in tree static void AddEdge( int u, int v)
{ // Edge from u to v
adj[u].add(v);
// Edge from V to U
adj[v].add(u);
} // Function that finds the maximum // matching of the DFS static void Matching_dfs( int u, int p)
{ for ( int i = 0 ; i < adj[u].size(); i++)
{
// Go further as we are not
// allowed to go towards
// its parent
if (adj[u].get(i) != p)
{
Matching_dfs(adj[u].get(i), u);
}
}
// If U and its parent P is
// not taken then we must
// take &mark them as taken
if (used[u] == 0 &&
used[p] == 0 && p != 0 )
{
// Increment size of edge set
max_matching++;
used[u] = used[p] = 1 ;
}
} // Function to find the maximum // matching in a graph static void maxMatching()
{ // Taking 1 as a root of the tree
Matching_dfs( 1 , 0 );
// Print maximum Matching
System.out.print(max_matching + "\n" );
} // Driver Code public static void main(String[] args)
{ for ( int i = 0 ; i < adj.length; i++)
adj[i] = new Vector<Integer>();
// Joining edge between
// two nodes in tree
AddEdge( 1 , 2 );
AddEdge( 1 , 3 );
AddEdge( 3 , 4 );
AddEdge( 3 , 5 );
// Function call
maxMatching();
} } // This code is contributed by amal kumar choubey |
# Python3 program for the above approach N = 10000
# Adjacency list to store edges adj = {}
used = [ 0 for i in range (N)]
max_matching = 0
# Add an edge between U and V in tree def AddEdge(u, v):
if u not in adj:
adj[u] = []
if v not in adj:
adj[v] = []
# Edge from u to v
adj[u].append(v)
# Edge from V to U
adj[v].append(u)
# Function that finds the maximum # matching of the DFS def Matching_dfs(u, p):
global max_matching
for i in range ( len (adj[u])):
# Go further as we are not
# allowed to go towards
# its parent
if (adj[u][i] ! = p):
Matching_dfs(adj[u][i], u)
# If U and its parent P is
# not taken then we must
# take &mark them as taken
if ( not used[u] and not used[p] and p ! = 0 ):
# Increment size of edge set
max_matching + = 1
used[u] = 1
used[p] = 1
# Function to find the maximum # matching in a graph def maxMatching():
# Taking 1 as a root of the tree
Matching_dfs( 1 , 0 )
# Print maximum Matching
print (max_matching)
# Driver Code n = 5
# Joining edge between # two nodes in tree AddEdge( 1 , 2 )
AddEdge( 1 , 3 )
AddEdge( 3 , 4 )
AddEdge( 3 , 5 )
# Function Call maxMatching() # This code is contributed by avanitrachhadiya2155 |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
static readonly int N = 10000;
// Adjacency list to store edges static List< int >[] adj = new List< int >[N];
static int []used = new int [N];
static int max_matching;
// Add an edge between U and V in tree static void AddEdge( int u, int v)
{ // Edge from u to v
adj[u].Add(v);
// Edge from V to U
adj[v].Add(u);
} // Function that finds the maximum // matching of the DFS static void Matching_dfs( int u, int p)
{ for ( int i = 0; i < adj[u].Count; i++)
{
// Go further as we are not
// allowed to go towards
// its parent
if (adj[u][i] != p)
{
Matching_dfs(adj[u][i], u);
}
}
// If U and its parent P is
// not taken then we must
// take &mark them as taken
if (used[u] == 0 &&
used[p] == 0 && p != 0)
{
// Increment size of edge set
max_matching++;
used[u] = used[p] = 1;
}
} // Function to find the maximum // matching in a graph static void maxMatching()
{ // Taking 1 as a root of the tree
Matching_dfs(1, 0);
// Print maximum Matching
Console.Write(max_matching + "\n" );
} // Driver Code public static void Main(String[] args)
{ for ( int i = 0; i < adj.Length; i++)
adj[i] = new List< int >();
// Joining edge between
// two nodes in tree
AddEdge(1, 2);
AddEdge(1, 3);
AddEdge(3, 4);
AddEdge(3, 5);
// Function call
maxMatching();
} } // This code is contributed by amal kumar choubey |
<script> // Javascript Program to implement the above approach
let N = 10000;
// Adjacency list to store edges
let adj = new Array(N);
let used = new Array(N);
used.fill(0);
let max_matching = 0;
// Add an edge between U and V in tree
function AddEdge(u, v)
{
// Edge from u to v
adj[u].push(v);
// Edge from V to U
adj[v].push(u);
}
// Function that finds the maximum
// matching of the DFS
function Matching_dfs(u, p)
{
for (let i = 0; i < adj[u].length; i++)
{
// Go further as we are not
// allowed to go towards
// its parent
if (adj[u][i] != p)
{
Matching_dfs(adj[u][i], u);
}
}
// If U and its parent P is
// not taken then we must
// take &mark them as taken
if (used[u] == 0 &&
used[p] == 0 && p != 0)
{
// Increment size of edge set
max_matching++;
used[u] = used[p] = 1;
}
}
// Function to find the maximum
// matching in a graph
function maxMatching()
{
// Taking 1 as a root of the tree
Matching_dfs(1, 0);
// Print maximum Matching
document.write(max_matching + "</br>" );
}
for (let i = 0; i < adj.length; i++)
adj[i] = [];
// Joining edge between
// two nodes in tree
AddEdge(1, 2);
AddEdge(1, 3);
AddEdge(3, 4);
AddEdge(3, 5);
// Function call
maxMatching();
</script> |
2
Time Complexity: O(V + E), where V is the number of edges and E is the number of edges.
Auxiliary Space: O(V)
Bottom-up DFS approach:
Another intuitive approach to solve this problem is to use DFS in a bottom-up manner and return two values at each level
Maximum matching including the current node
Maximum matching excluding the current node
We will recurse on the left and right subtrees and get these values for both of them. We can then calculate new values for the current level based on these values.
Let left_included denote maximum matching including root of left subtree and left_excluded denote the maximum matching excluding the root of left subtree. Similarly, for right_included and right_excluded.
If we include the current node in maximum matching, then we have to exclude one of either left sub tree root or right sub tree root. Including both will cause overlapping on current node which is not allowed. By excluding either left or right sub tree root we can increase the maximum matching by 1 by including one of the edges from current_node -> left sub tree root or current_node -> right sub tree root.
Thus maximum matching including current node will be given by
current_including = max(max(left_including, right_excluding) + 1, max(left_excluding, right_including) + 1)
If we exclude the current node then we can include both left and right subtree root. As matchings in left and right subtrees are independent of each other we can get maximum value by adding both matchings.
Thus maximum matching excluding current node will be given by
current_excluding = left_including + right_including
We will return both these values from the current recursion level to upper recursion levels. After the recursion completes we will receive two values, maximum matching including root node and maximum matching excluding root node.
The maximum of those two will give the maximum matching in the tree.
#include <iostream> #include <algorithm> using namespace std;
class Node {
public :
Node* left;
Node* right;
int val;
Node( int key) {
this ->left = nullptr;
this ->right = nullptr;
this ->val = key;
}
}; int * maxMatchingHelper(Node* root) {
if (root == nullptr) {
int * arr = new int [2];
arr[0] = 0;
arr[1] = 0;
return arr;
}
if (root->left == nullptr && root->right == nullptr) {
int * arr = new int [2];
arr[0] = 0;
arr[1] = 0;
return arr;
}
int * left = maxMatchingHelper(root->left);
int * right = maxMatchingHelper(root->right);
// Maximum matching including current node
int currIncluded = max(max(left[0], right[1]) + 1, max(left[1], right[0]) + 1);
// Maximum matching excluding current node
int currExcluded = left[0] + right[0];
return new int [2] { currIncluded, currExcluded };
} int maxMatching(Node* root) {
// Taking 1 as a root of the tree
int * result = maxMatchingHelper(root);
// Return maximum Matching
return max(result[0], result[1]);
} int main() {
// Taking 1 as a root of the tree
Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(7);
root->left->left = new Node(3);
root->left->right = new Node(4);
root->left->right->left = new Node(5);
root->left->right->right = new Node(6);
root->right->left = new Node(8);
root->right->right = new Node(9);
cout << maxMatching(root) << endl;
return 0;
} |
class Node {
Node left, right;
int val;
public Node( int key) {
this .left = null ;
this .right = null ;
this .val = key;
}
} public class Main {
public static void main(String[] args) {
// Taking 1 as a root of the tree
Node root = new Node( 1 );
root.left = new Node( 2 );
root.right = new Node( 7 );
root.left.left = new Node( 3 );
root.left.right = new Node( 4 );
root.left.right.left = new Node( 5 );
root.left.right.right = new Node( 6 );
root.right.left = new Node( 8 );
root.right.right = new Node( 9 );
System.out.println(maxMatching(root));
}
public static int [] maxMatchingHelper(Node root) {
if (root == null ) {
return new int []{ 0 , 0 };
}
if (root.left == null && root.right == null ) {
return new int []{ 0 , 0 };
}
int [] left = maxMatchingHelper(root.left);
int [] right = maxMatchingHelper(root.right);
// Maximum matching including current node
int currIncluded = Math.max(
Math.max(left[ 0 ], right[ 1 ]) + 1 ,
Math.max(left[ 1 ], right[ 0 ]) + 1
);
// Maximum matching excluding current node
int currExcluded = left[ 0 ] + right[ 0 ];
return new int []{currIncluded, currExcluded};
}
public static int maxMatching(Node root) {
// Taking 1 as a root of the tree
int [] result = maxMatchingHelper(root);
// Return maximum Matching
return Math.max(result[ 0 ], result[ 1 ]);
}
} |
class Node:
def __init__( self , key):
self .left = None
self .right = None
self .val = key
def max_matching_helper(root):
if not root:
return ( 0 , 0 )
if not root.left and not root.right:
return ( 0 , 0 )
left_included, left_excluded = max_matching_helper(root.left)
right_included, right_excluded = max_matching_helper(root.right)
# Maximum matching gincluding current node
curr_included = max ( max (left_included, right_excluded) + 1 , max (left_excluded, right_included) + 1 )
# Maximum matching excluding current node
curr_excluded = left_included + right_included
return (curr_included, curr_excluded)
def max_matching(root):
# Taking 1 as a root of the tree
root_including, root_excluding = max_matching_helper(root)
# Return maximum Matching
return max (root_including, root_excluding)
# Driver code root = Node( 1 )
root.left = Node( 2 )
root.right = Node( 7 )
root.left.left = Node( 3 )
root.left.right = Node( 4 )
root.left.right.left = Node( 5 )
root.left.right.right = Node( 6 )
root.right.left = Node( 8 )
root.right.right = Node( 9 )
print (max_matching(root))
# This code is contributed by Rathijeet Bhave |
using System;
public class Node {
public Node left;
public Node right;
public int val;
public Node( int key) {
this .left = null ;
this .right = null ;
this .val = key;
}
} public class MaximumMatching {
// Helper function to compute the maximum matching for a node in the tree
private static int [] MaxMatchingHelper(Node root) {
if (root == null ) {
int [] arr = new int [2];
arr[0] = 0;
arr[1] = 0;
return arr;
}
if (root.left == null && root.right == null ) {
int [] arr = new int [2];
arr[0] = 0;
arr[1] = 0;
return arr;
}
int [] left = MaxMatchingHelper(root.left);
int [] right = MaxMatchingHelper(root.right);
// Maximum matching including current node
int currIncluded = Math.Max(Math.Max(left[0], right[1]) + 1, Math.Max(left[1], right[0]) + 1);
// Maximum matching excluding current node
int currExcluded = left[0] + right[0];
return new int [2] { currIncluded, currExcluded };
}
// Function to compute the maximum matching for a binary tree
public static int MaxMatching(Node root) {
// Taking 1 as the root of the tree
int [] result = MaxMatchingHelper(root);
// Return maximum matching
return Math.Max(result[0], result[1]);
}
// Main function to test the implementation
public static void Main() {
// Taking 1 as the root of the tree
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(7);
root.left.left = new Node(3);
root.left.right = new Node(4);
root.left.right.left = new Node(5);
root.left.right.right = new Node(6);
root.right.left = new Node(8);
root.right.right = new Node(9);
Console.WriteLine(MaxMatching(root)); // Output: 5
}
} |
// JavaScript code for the above approach
class Node {
constructor(key) {
this .left = null ;
this .right = null ;
this .val = key;
}
}
function maxMatchingHelper(root) {
if (!root) {
return [0, 0];
}
if (!root.left && !root.right) {
return [0, 0];
}
const [leftIncluded, leftExcluded] = maxMatchingHelper(root.left);
const [rightIncluded, rightExcluded] = maxMatchingHelper(root.right);
// Maximum matching including current node
const currIncluded = Math.max(
Math.max(leftIncluded, rightExcluded) + 1,
Math.max(leftExcluded, rightIncluded) + 1
);
// Maximum matching excluding current node
const currExcluded = leftIncluded + rightIncluded;
return [currIncluded, currExcluded];
}
function maxMatching(root)
{
// Taking 1 as a root of the tree
const [rootIncluding, rootExcluding] = maxMatchingHelper(root);
// Return maximum Matching
return Math.max(rootIncluding, rootExcluding);
}
// Driver code
const root = new Node(1);
root.left = new Node(2);
root.right = new Node(7);
root.left.left = new Node(3);
root.left.right = new Node(4);
root.left.right.left = new Node(5);
root.left.right.right = new Node(6);
root.right.left = new Node(8);
root.right.right = new Node(9);
console.log(maxMatching(root));
// This code is contributed by Potta Lokesh. |
3
Time Complexity: O(V + E), where V is the number of edges and E is the number of edges.
Auxiliary Space: O(V)