[GFG] leftmost and rightmost nodes of binary tree

// AMAN JAIN MCA 1st YEAR 2nd SEM

// time O(N), space O(number of nodes in a level)
// Approach: Basic level order traversal using level separators
void printCorner(Node *root) {
    vector<int> currentLevel;
    queue<Node*> nodeQueue;
    nodeQueue.push(root);
    nodeQueue.push(NULL);
    while(!nodeQueue.empty()) {
        Node* currentNode = nodeQueue.front();
        nodeQueue.pop();
        if(currentNode == NULL) {
            cout << currentLevel[0] << " ";
            if(currentLevel.size() > 1) {
                cout << currentLevel[currentLevel.size() - 1] << " ";
            }
            currentLevel = vector<int>();
            if(nodeQueue.size()) {
                nodeQueue.push(NULL);
            }
        }
        else {
            currentLevel.push_back(currentNode -> data);
            if(currentNode -> left) {
                nodeQueue.push(currentNode -> left);
            }
            if(currentNode -> right) {
                nodeQueue.push(currentNode -> right);
            }
        }
    }
}