Find maximum matching in a given Binary Tree - GeeksforGeeks

23 Feb, 2022
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.
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++
Java
Python3
C#
Javascript
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;#define N 10000 // Adjacency list to store edgesvector<int> adj[N]; int used[N];int max_matching; // Add an edge between U and V in treevoid 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 DFSvoid 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 graphvoid maxMatching(){    // Taking 1 as a root of the tree    Matching_dfs(1, 0);     // Print maximum Matching    cout << max_matching << "\n";} // Driver Codeint 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 approachimport 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 treestatic 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 DFSstatic 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 graphstatic void maxMatching(){         // Taking 1 as a root of the tree    Matching_dfs(1, 0);     // Print maximum Matching    System.out.print(max_matching + "\n");} // Driver Codepublic 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 approachN = 10000 # Adjacency list to store edgesadj = {} used = [0 for i in range(N)] max_matching = 0 # Add an edge between U and V in treedef 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 DFSdef 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 graphdef maxMatching():     # Taking 1 as a root of the tree    Matching_dfs(1, 0)     # Print maximum Matching    print(max_matching)     # Driver Coden = 5 # Joining edge between# two nodes in treeAddEdge(1, 2)AddEdge(1, 3)AddEdge(3, 4)AddEdge(3, 5) # Function CallmaxMatching() # This code is contributed by avanitrachhadiya2155
// C# program for the above approachusing System;using System.Collections.Generic; class GFG{     static readonly int N = 10000; // Adjacency list to store edgesstatic 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 treestatic 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 DFSstatic 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 graphstatic void maxMatching(){         // Taking 1 as a root of the tree    Matching_dfs(1, 0);     // Print maximum Matching    Console.Write(max_matching + "\n");} // Driver Codepublic 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.
Python3
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 coderoot = 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
3
 Time Complexity: O(V + E), where V is the number of edges and E is the number of edges.
 Auxiliary Space: O(V)
Amal Kumar Choubey
avanitrachhadiya2155
rathijeetbhave
divyesh072019
sumitgumber28
DFS
PostOrder Traversal
Tree Traversals
tree-traversal
Trees
Greedy
Tree
Greedy
DFS
Tree
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