Level Order Traversal of a Binary Tree

void traverse(Node* root)
{
    queue<Node*> q;

    if (root) {
        q.push(root);
    }
    while (!q.empty())
    {
        const Node * const temp_node = q.front();
        q.pop();
        cout<<temp_node->value<<"\n";

        if (temp_node->left) {
            q.push(temp_node->left);
        }
        if (temp_node->right) {
            q.push(temp_node->right);
        }
    }
}

There, no more special case. And the indentation is cleaned up so it can be understood more easily.

Alternatively:

void traverse(Node* root)
{
    queue<Node*> q;

    if (!root) {
        return;
    }
    for (q.push(root); !q.empty(); q.pop()) {
        const Node * const temp_node = q.front();
        cout<<temp_node->value<<"\n";

        if (temp_node->left) {
            q.push(temp_node->left);
        }
        if (temp_node->right) {
            q.push(temp_node->right);
        }
    }
}

Done up as a for loop. Personally, I like the extra variable. The variable name is a nicer shorthand than saying 'q.front()` all the time.

You can try this way:

struct Node
{
    char data;
    Node* left;
    Node* right;
};
void LevelOrder(Node* root)
{
    if(root == NULL) return;
    queue<Node*> Q;
    Q.push(root);
    while(!Q.empty())
    {
        Node* current = Q.front();
        cout<< current->data << " ";
        if(current->left != NULL) Q.push(current->left);
        if(current->right != NULL) Q.push(current->right);
        Q.pop();
    }
}

One serious problem with your existing code is it crashes when it is called on an empty tree (root = NULL).

You need to decide if you want to have NULL pointers in the queue or not.

If not them you can only enqueue non-NULL values.

void traverse(Node* root) {
    queue<Node*> q;

    // no tree no level order.
    if(root == NULL) {
        return;
    }

    // push the root to start with as we know it is not NULL.
    q.push(root);

    // loop till there are nodes in the queue.
    while(!q.empty()) {
        // dequeue the front node.
        Node *tmpNode = q.front();
        q.pop();

        // print it..we are sure it is not NULL.
        cout<<tmpNode->value<<" ";

        // enqueue left child if it exists.
        if(tmpNode->left) {
            q.push(tmpNode->left);
        }
        // enqueue right child if it exists.
        if(tmpNode->right) {
            q.push(tmpNode->right);
        }
    }
}

Alternatively if you decide to have NULL in the queue you can do:

void traverse(Node* root) {
    queue<Node*> q;

    // push the root..even if it is NULL.
    q.push(root);

    // loop till the queue is not empty.
    while(!q.empty()) {
        // dequeue the front node.
        Node *tmpNode = q.front();
        q.pop();

        // the dequeued pointer can be NULL or can point to a node.
        // process the node only if it is not NULL.     
        if(tmpNode) {       
            cout<<tmpNode->value<<" ";
            q.push(tmpNode->left);
            q.push(tmpNode->right);
        }
    }   
}

The first method is preferred as a large tree has plenty of NULL children (children of leaf nodes) and there is no point in having them enqueued in the queue when we later just don't process them.