Given two queues with their standard operations (enqueue, dequeue, isempty, size), implement a stack with its standard operations (pop, push, isempty, size).
To solve this problem first we have to declare one queue as working queue then enqueue the first element in to that queue. after that when we have to push another element then we push another element in other queue then dequeue all elements of working queue one by one till it is empty and enqueue all into other queue in same order and set other queue as working queue. this is for pushing elements into stack(implemented in queues). for pop operation we just have to dequeue element from working queue that will be the top element for the stack. isempty = same isempty operation on working queue. size= size of the working queue.
Given a Binary Tree, find vertical sum of the nodes that are in same vertical line. Print all sums through different vertical lines. Examples: 1 / \ 2 3 / \ / \ 4 5 6 7 The tree has 5 vertical lines Vertical-Line-1 has only one node 4 => vertical sum is 4 Vertical-Line-2: has only one node 2=> vertical sum is 2 Vertical-Line-3: has three nodes: 1,5,6 => vertical sum is 1+5+6 = 12 Vertical-Line-4: has only one node 3 => vertical sum is 3 Vertical-Line-5: has only one node 7 => vertical sum is 7 So expected output is 4, 2, 12, 3 and 7
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ReplyDeleteTo solve this problem first we have to declare one queue as working queue then enqueue the first element in to that queue.
ReplyDeleteafter that when we have to push another element then we push another element in other queue then dequeue all elements of working queue one by one till it is empty and enqueue all into other queue in same order and set other queue as working queue.
this is for pushing elements into stack(implemented in queues).
for pop operation we just have to dequeue element from working queue that will be the top element for the stack.
isempty = same isempty operation on working queue.
size= size of the working queue.
@gaurav yes this will work fine..:)
ReplyDelete