Find maximum matching in a given Binary Tree

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: 

1. 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.
2. 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.
3. 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)