# Check whether given degrees of vertices represent a Graph or Tree

Given the number of vertices and the degree of each vertex where vertex numbers are 1, 2, 3,…n. The task is to identify whether it is a graph or a tree. It may be assumed that the graph is Connected.

Examples:

```Input : 5
2 3 1 1 1
Output : Tree
Explanation : The input array indicates that
vertex one has degree 2, vertex
two has degree 3, vertices 3, 4
and 5 have degree 1.
1
/ \
2   3
/ \
4   5

Input : 3
2 2 2
Output : Graph
1
/ \
2 - 3```

The degree of a vertex is given by the number of edges incident or leaving from it. This can simply be done using the properties of trees like –

1. Tree is connected and has no cycles while graphs can have cycles.
2. Tree has exactly n-1 edges while there is no such constraint for graph.
3. It is given that the input graph is connected. We need at least n-1 edges to connect n nodes.

If we take the sum of all the degrees, each edge will be counted twice. Hence, for a tree with n vertices and n – 1 edges, sum of all degrees should be 2 * (n – 1). Please refer Handshaking Lemma for details. So basically we need to check if sum of all degrees is 2*(n-1) or not.

Implementation:

## C++

 `// C++ program to check whether input degree` `// sequence is graph or tree` `#include` `using` `namespace` `std;`   `// Function returns true for tree` `// false for graph` `bool` `check(``int` `degree[], ``int` `n)` `{` `    ``// Find sum of all degrees` `    ``int` `deg_sum = 0;` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``deg_sum += degree[i];`   `    ``// Graph is tree if sum is equal to 2(n-1)` `    ``return` `(2*(n-1) == deg_sum);` `}`   `// Driver program to test above function` `int` `main()` `{` `    ``int` `n = 5;` `    ``int` `degree[] = {2, 3, 1, 1, 1};`   `    ``if` `(check(degree, n))` `        ``cout << ``"Tree"``;` `    ``else` `        ``cout << ``"Graph"``;`   `    ``return` `0;` `}`

## Java

 `// Java program to check whether input degree ` `// sequence is graph or tree ` `class` `GFG ` `{`   `    ``// Function returns true for tree ` `    ``// false for graph ` `    ``static` `boolean` `check(``int` `degree[], ``int` `n)` `    ``{` `        ``// Find sum of all degrees ` `        ``int` `deg_sum = ``0``;` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{` `            ``deg_sum += degree[i];` `        ``}`   `        ``// Graph is tree if sum is equal to 2(n-1) ` `        ``return` `(``2` `* (n - ``1``) == deg_sum);` `    ``}`   `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `n = ``5``;` `        ``int` `degree[] = {``2``, ``3``, ``1``, ``1``, ``1``};`   `        ``if` `(check(degree, n))` `        ``{` `            ``System.out.println(``"Tree"``);` `        ``} ` `        ``else` `        ``{` `            ``System.out.println(``"Graph"``);` `        ``}` `    ``}` `} `     `// This code has been contributed ` `// by 29AjayKumar`

## Python

 `# Python program to check whether input degree` `# sequence is graph or tree` `def` `check(degree, n):` `    `  `    ``# Find sum of all degrees` `    ``deg_sum ``=` `sum``(degree)` `    `  `    ``# It is tree if sum is equal to 2(n-1)` `    ``if` `(``2``*``(n``-``1``) ``=``=` `deg_sum):` `        ``return` `True` `    ``else``:` `        ``return` `False` `    `  `#main` `n ``=` `5` `degree ``=` `[``2``, ``3``, ``1``, ``1``, ``1``];` `if` `(check(degree, n)):` `    ``print` `"Tree"` `else``:` `    ``print` `"Graph"`

## C#

 `// C# program to check whether input ` `// degree sequence is graph or tree ` `using` `System;`   `class` `GFG ` `{`   `    ``// Function returns true for tree ` `    ``// false for graph ` `    ``static` `bool` `check(``int``[] degree, ``int` `n)` `    ``{` `        ``// Find sum of all degrees ` `        ``int` `deg_sum = 0;` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{` `            ``deg_sum += degree[i];` `        ``}`   `        ``// Graph is tree if sum is ` `        ``// equal to 2(n-1) ` `        ``return` `(2 * (n - 1) == deg_sum);` `    ``}`   `    ``// Driver code ` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `n = 5;` `        ``int``[] degree = {2, 3, 1, 1, 1};`   `        ``if` `(check(degree, n))` `        ``{` `            ``Console.WriteLine(``"Tree"``);` `        ``} ` `        ``else` `        ``{` `            ``Console.WriteLine(``"Graph"``);` `        ``}` `    ``}` `} `   `// This code has been contributed ` `// by Akanksha Rai`

## PHP

 ``

## Javascript

 `// JS program to check whether input degree ` `// sequence is graph or tree `   ` ``// Function returns true for tree ` `// false for graph ` `function` `check(degree, n)` `{`   `  ``// Find sum of all degrees ` `  ``const deg_sum = degree.reduce((a, b) => a + b, 0);` `  `  `  ``// Graph is tree if sum is equal to 2(n-1) ` `  ``if` `(2 * (n - 1) === deg_sum) {` `    ``return` `true``;` `  ``} ``else` `{` `    ``return` `false``;` `  ``}` `}`   `// Driver code` `const n = 5;` `const degree = [2, 3, 1, 1, 1];` `if` `(check(degree, n)) {` ` ``console.log(``"Tree"``);` `} ``else` `{` ` ``console.log(``"Graph"``);` `}`

Output

`Tree`

Time Complexity:O(N)

Space Complexity:O(1),since no extra space being used.

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