You are given an undirected graph of N nodes and M edges. Return 1 if the graph is a tree, else return 0.

To solve this problem, you need to understand the properties of a tree in graph theory. A graph is a tree if it is connected and does not contain any cycle.

Here are the steps to solve this problem:

1. Initialize a data structure (like an adjacency list) to store the graph.

2. Check if the graph is connected. You can do this by performing a Depth-First Search (DFS) or Breadth-First Search (BFS) from any node and then checking if all nodes were visited. If not all nodes were visited, the graph is not connected and you can return 0.

3. Check if the graph contains a cycle. You can do this by performing a DFS and checking if you ever visit a node that has already been visited. If you find such a node, the graph contains a cycle and you can return 0.

4. If the graph is connected and does not contain a cycle, it is a tree. Return 1.

Here is a Python code snippet that implements these steps:

def isTree(N, M, edges):
    # Step 1: Store the graph
    graph = [[] for _ in range(N+1)]
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    # Step 2: Check if the graph is connected
    visited = [0]*(N+1)
    def dfs(node, parent):
        visited[node] = 1
        for child in graph[node]:
            if visited[child] == 0:
                if dfs(child, node) == False:
                    return False
            elif child != parent:
                return False
        return True

    # If the graph is not connected
    if dfs(1, -1) == False or sum(visited) != N:
        return 0

    # Step 3: Check if the graph contains a cycle
    # If the graph is connected and does not contain a cycle
    return 1

This function takes as input the number of nodes N, the number of edges M, and a list of edges, where each edge is a tuple of two nodes. It returns 1 if the graph is a tree and 0 otherwise.