Print right view of the tree

Given a binary search tree, print right view of it, i.e all the nodes which will be seen if binary search tree is looked at from right hand side
http://www.geeksforgeeks.org/print-right-view-binary-tree-2/

The key is: Pre-Order Traverse, but right child first: Root -> Right -> Left
The Right view contains all nodes that are last nodes in their levels. A simple solution is to dolevel order traversal and print the last node in every level.

The problem can also be solved using simple recursive traversal. We can keep track of level of a node by passing a parameter to all recursive calls. The idea is to keep track of maximum level also. And traverse the tree in a manner that right subtree is visited before left subtree. Whenever we see a node whose level is more than maximum level so far, we print the node because this is the last node in its level (Note that we traverse the right subtree before left subtree).

void rightViewUtil(struct Node *root, int level, int *max_level)

{

    // Base Case

    if (root==NULL)  return;

    // If this is the last Node of its level

    if (*max_level < level)

    {

        printf("%d\t", root->data);

        *max_level = level;

    }

    // Recur for right subtree first, then left subtree

    rightViewUtil(root->right, level+1, max_level);

    rightViewUtil(root->left, level+1, max_level);

}

void rightView(struct Node *root)

{

    int max_level = 0;

    rightViewUtil(root, 1, &max_level);

}

https://gyaneshwarpardhi.wordpress.com/2014/06/22/print-right-view-of-a-binary-tree/
BFS, Level Order Traverse => Using two queues(Or use a count)

static void printRightViewOfBinaryTree(treeNode *rootNode) {

    //queues to store current and next level nodes

    queue<treeNode*> Q1;

    queue<treeNode*> Q2;

    Q1.push(rootNode);

    while (!Q1.empty()) {

        treeNode * temp = Q1.front();

        Q1.pop();

        if (temp->leftChild)

            Q2.push(temp->leftChild);

        if (temp->rightChild)

            Q2.push(temp->rightChild);

        if (Q1.empty()) {

            //print last element of each level

            cout << temp->val << endl;

            //move nodes form next level to current level

            swap(Q1, Q2);

        }

    }

}

Alternative solution: Use an queue to do level traversal and Print last value in each level.

The Right view contains all nodes that are last nodes in their levels. A simple solution is to do level order traversal and print the last node in every level.

Pre-order traversal, but right tree first then left tree, also keep track of maximum level so far.

void print_right_view(Node * node, int curr_level, int *max_level){

if(node == NULL) return;

if(curr_level >  *max_level){

printf("%d  ", node->value);

*max_level = curr_level;

}

print_right_view(node->right, curr_level+1, max_level);

print_right_view(node->left, curr_level+1, max_level);

}

Read full article from Algorithms and Me: Binary Search Tree : Print right view of the tree