A tree is always a Bipartite Graph as we can always break into two disjoint sets with alternate levels. In other words we always color it with two colors such that alternate levels have same color. The task is to compute the maximum no. of edges that can be added to the tree so that it remains Bipartite Graph.

**Examples:**

Input : Tree edges as vertex pairs 1 2 1 3 Output : 0 Explanation : The only edge we can add is from node 2 to 3. But edge 2, 3 will result in odd cycle, hence violation of Bipartite Graph property. Input : Tree edges as vertex pairs 1 2 1 3 2 4 3 5 Output : 2 Explanation : On colouring the graph, {1, 4, 5} and {2, 3} form two different sets. Since, 1 is connected from both 2 and 3, we are left with edges 4 and 5. Since, 4 is already connected to 2 and 5 to 3, only options remain {4, 3} and {5, 2}.

1) Do a simple DFS (or BFS) traversal of graph and color it with two colors.

2) While coloring also keep track of counts of nodes colored with the two colors. Let the two counts be count_color_{0} and count_color_{1}.

3) Now we know maximum edges a bipartite graph can have are count_color_{0} x count_color_{1}.

4) We also know tree with n nodes has n-1 edges.

5) So our answer is count_color_{0} x count_color_{1} – (n-1).

Below is the implementation :

## C++

`// CPP code to print maximum edges such that ` `// Tree remains a Bipartite graph ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// To store counts of nodes with two colors ` `long` `long` `count_color[2]; ` ` ` `void` `dfs(vector<` `int` `> adj[], ` `int` `node, ` `int` `parent, ` `int` `color) ` `{ ` ` ` `// Increment count of nodes with current ` ` ` `// color ` ` ` `count_color[color]++; ` ` ` ` ` `// Traversing adjacent nodes ` ` ` `for` `(` `int` `i = 0; i < adj[node].size(); i++) { ` ` ` ` ` `// Not recurring for the parent node ` ` ` `if` `(adj[node][i] != parent) ` ` ` `dfs(adj, adj[node][i], node, !color); ` ` ` `} ` `} ` ` ` `// Finds maximum number of edges that can be added ` `// without violating Bipartite property. ` `int` `findMaxEdges(vector<` `int` `> adj[], ` `int` `n) ` `{ ` ` ` `// Do a DFS to count number of nodes ` ` ` `// of each color ` ` ` `dfs(adj, 1, 0, 0); ` ` ` ` ` `return` `count_color[0] * count_color[1] - (n - 1); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 5; ` ` ` `vector<` `int` `> adj[n + 1]; ` ` ` `adj[1].push_back(2); ` ` ` `adj[1].push_back(3); ` ` ` `adj[2].push_back(4); ` ` ` `adj[3].push_back(5); ` ` ` `cout << findMaxEdges(adj, n); ` ` ` `return` `0; ` `} ` |

## Python3

# Python 3 code to print maximum edges such

# that Tree remains a Bipartite graph

def dfs(adj, node, parent, color):

# Increment count of nodes with

# current color

count_color[color] += 1

# Traversing adjacent nodes

for i in range(len(adj[node])):

# Not recurring for the parent node

if (adj[node][i] != parent):

dfs(adj, adj[node][i],

node, not color)

# Finds maximum number of edges that

# can be added without violating

# Bipartite property.

def findMaxEdges(adj, n):

# Do a DFS to count number of

# nodes of each color

dfs(adj, 1, 0, 0)

return (count_color[0] *

count_color[1] – (n – 1))

# Driver code

# To store counts of nodes with

# two colors

count_color = [0, 0]

n = 5

adj = [[] for i in range(n + 1)]

adj[1].append(2)

adj[1].append(3)

adj[2].append(4)

adj[3].append(5)

print(findMaxEdges(adj, n))

# This code is contributed by PranchalK

**Output:**

2

**Time Complexity:** O(n)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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