Groupon Interview Question for Software Engineer in Tests


Country: India
Interview Type: Phone Interview




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2
of 4 vote

use list as a queue (maintain a pointer to its tail)
put the root node in the queue
while there are nodes in the queue:
- pop a node from the queue
- print it
- push its children to the tail of the queue

basically, it's BFS

- gen-y-s December 03, 2012 | Flag Reply
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0
of 0 votes

That will print: x, xl, xr, xll, xlr, xrl, xrr

- glazmo December 08, 2012 | Flag
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0
of 0 votes

level order traversal it is.

- nishant.cena April 23, 2013 | Flag
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0
of 0 vote

x
/ \
xl xr
/ \ / \
xll xlr xrl xrr

then tree should be like...
---xll
---xl
---x-xlr-xrl
---xr
---xrr

- Anonymous December 02, 2012 | Flag Reply
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0
of 0 votes

Couldn't understand the order

- dinnu.k December 03, 2012 | Flag
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0
of 0 votes

Assign column numbers to the nodes first...
assign root as 0, then immediate left node as -1 and right node as +1..now proceed to next level...assign -1 to left tree node and assign +1 to right tree node...now...the xlr and xrl have same column number, that is 0..hence same with the root..this is how we can print x-xlr-xrl..coz they have the same column number,,,((-1)+(+1)=0)..all left tree will have -1 and -2 as their column number now...and all right tree have +1 and +2 as their column number now...try it...pls drop ur comments if you have any other logic..

- Anonymous December 03, 2012 | Flag
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0
of 0 vote

Algorithm: Do a BFS and keep the nodes in an array. So, for any index of the array i, the vertical order match would be in the relation 4i+1 and 4i+2. Use recursion to print all the nodes for a given i. For ex. for i = 1, the vertical nodes are 5, 6, (4*5+1),(4*5 +1),(4*6 +1), (4*6+2)th node of the array and so on. Print those and make those nodes infinity in the array, so that they are not printed again. Then iterate through the whole array.

- Anonymous January 18, 2013 | Flag Reply
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0
of 0 vote

Do lever order Traversal :

www [dot] geeksforgeeks [dot] org/level-order-tree-traversal

- Veera January 22, 2013 | Flag Reply
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0
of 0 vote

Do Level Order Traversal :

www[dot]geeksforgeeks[dot]org/level-order-tree-traversal/

- Veera January 22, 2013 | Flag Reply
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0
of 0 vote

public void printVerticalTraversal() {
		Map<Integer,LinkedList<TreeNode>> hashtable = new HashMap<>();
		printVerticalTraversalHelper(root,0,hashtable);
		printHashTable(hashtable);
	}

	private void printHashTable(Map<Integer, LinkedList<TreeNode>> hashtable) {
		for (Integer key : hashtable.keySet()) {
			LinkedList<TreeNode> list = hashtable.get(key);
			System.out.println("Key = "+key);
			printList(list);
		}
	}

	private void printList(LinkedList<TreeNode> list) {
		for (TreeNode node : list) {
			System.out.print(node.getData());
			System.out.print(" ");
		}
		System.out.println("\n");
	}

	private void printVerticalTraversalHelper(TreeNode root, int value, Map<Integer, LinkedList<TreeNode>> hashtable) {
		addToHashTable(value,root,hashtable);
		if(root.getLeft()!=null)
			printVerticalTraversalHelper(root.getLeft(), value-1, hashtable);
		if(root.getRight()!=null)
			printVerticalTraversalHelper(root.getRight(), value+1, hashtable);
	}

	private void addToHashTable(int value, TreeNode root,
			Map<Integer, LinkedList<TreeNode>> hashtable) {
		LinkedList<TreeNode> result = new LinkedList<>();
		if(hashtable.containsKey(value))
			result = hashtable.get(value);
		result.add(root);
		hashtable.put(value, result);
	}

- rvm December 07, 2013 | Flag Reply
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0
of 0 vote

@rvm
This wont work. Java is pass by value.

- sammy October 04, 2014 | Flag Reply


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