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.

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 –

- Tree is
**connected**and has**no cycles**while graphs can have cycles. - Tree has exactly
**n-1**edges while there is no such constraint for graph. - 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) ore not.

## C++

`// C++ program to check whether input degree ` `// sequence is graph or tree ` `#include<bits/stdc++.h> ` `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; ` `} ` |

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## 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"` |

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**Output:**

Tree

This article is contributed by **Ayush Khanduri**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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