CareerCup

Look at below link for solution...

http://tech-queries.blogspot.com/2008/12/level-order-tree-traversal-in-reverse.html

q.push_back(root)

while (q is not empty) {
node = q.pop_front();
s.push(node)'

if (node->left) {
q.push_back(node->left)
}
if (node->right) {
q.push_back(node->right)
}
}

while (S !empty) {
pop and print.
}

algorithm

(1) while simple traverse make all node's left pointer pointing to its parent, when u get leaves just push it in queue with its level

(2) while considering the levels of the nodes in the queue print the node values

(3) if required revert back to the original tree

1. Start from root and enqueue to the queue.
2. Dequeue an element and push it to the stack.
3. Enqueue the children from the dequeued node(from step 2).
4. Repeat step 2 and 3 until you reach the leaves.
5. Now print the stack by popping elements until it gets empty.

Hope it's correct! :)

Use "reverse-dfs" to solve the problem.
@ Akash, where did you copy those code chunks from?seems to be from some data structure book. anyways that does the same thing as I mentioned above.

use BFS. it traverse all the node level wise and it requires only one queue.