Google Interview Question for SDE1s


Country: United States




Comment hidden because of low score. Click to expand.
0
of 0 vote

Solution for the first part.

Follow-up: Yes, it will be simpler to solve, since searching for the node in the forest will be faster. In the case of the BST, only the find_node function needs to be changed in the below implementation, everything else remains the same.

class TreeNode():
    def __init__(self, x, left = None, right = None):
        self.val = x
        self.left = left
        self.right = right

    def print_tree(self, level=0):
        print(level*"  " + str(self.val))
        if self.left:
            self.left.print_tree(level+1)
        if self.right:
            self.right.print_tree(level+1)


def T(x, l=None, r=None):
    """Convenience function for TreeNodes"""
    return TreeNode(x, l, r)

def main():
    print("TEST 1: Erasing [1, 5]")
    test1()
    print("TEST 2: Erasing [2, 3]")
    test2()

def test1():
    #      1
    #   2     3
    # 4  5   6  7
    root = T(1, T(2, T(4), T(5)), T(3, T(6), T(7)))
    to_be_erased = [1, 5]
    forest = break_tree(root, to_be_erased)
    print_forest(forest)

def test2():
    #      1
    #   2     3
    # 4  5   6  7
    root = T(1, T(2, T(4), T(5)), T(3, T(6), T(7)))
    to_be_erased = [2, 3]
    forest = break_tree(root, to_be_erased)
    print_forest(forest)

def print_forest(forest):
    for i, f in enumerate(forest):
        print("Forest #" + str(i + 1))
        f.print_tree()
        print()

def break_tree(root, to_be_erased):
    forest = [root]

    for num in to_be_erased:
        for i, rt in enumerate(forest):
            n, parent, which = find_node(rt, num)
            if n:
                forest.append(n.left)
                forest.append(n.right)

                # If this node has no parent, it means it was a root in the forest
                # and we need to remove it from our list
                if not parent:
                    forest[i] = None
                else:
                    if which == "left":
                        parent.left = None
                    elif which == "right":
                        parent.right = None

    return [f for f in forest if f]


def find_node(root, num):
    def find_node_and_parent(root, num, parent, which):
        if root:
            if root.val == num:
                return root, parent, which
            else:
                n, p, w = find_node_and_parent(root.left, num, root, "left")
                if n:
                    return n, p, w
                n, p, w = find_node_and_parent(root.right, num, root, "right")
                if n:
                    return n, p, w

        return None, None, None

    return find_node_and_parent(root, num, None, None)


if __name__ == "__main__":
    main()

Output:

TEST 1: Erasing [1, 5]
Forest #1
2
  4

Forest #2
3
  6
  7

TEST 2: Erasing [2, 3]
Forest #1
1

Forest #2
4

Forest #3
5

Forest #4
6

Forest #5
7

- havanagrawal November 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Traverse the tree in pre order, so any node is erased before it's children. Here is the high level algorithm.

isRootErased := false
forest := []

eraseNodes(node, parent):
  if node is null
    return

  if shouldBeErased(node):
    if parent is null:
      isRootErased := true
    else if parent.left is node:
      parent.left := null
    else:
      parent.right := null

    if node.left is not null:
      forest.push(node.left)

    if node.right is not null:
      forest.push(node.right)

  eraseNodes(node.left, node)
  eraseNodes(node.right, node)

eraseNodes(root, null)

if not isRootErased:
  forest.push(root)

return forest

Processing each node take O(1), so time complexity is O(N), where N is the number of nodes.

Regarding the follow up, it depends. Suppose you are given a list of K nodes to erase. Finding each of them would take O(log N) if the tree is balanced. So overall complexity in time would be O(K log N), which is better than O(N) if K is small. In the worst case, K = N and we get O(N log N), which is worse.

- Anonymous November 14, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class node
{
public:
	int val;
	node* left;
	node* right;
	node( int v): val(v), left( nullptr), right(nullptr)
	{}
};

node* eraseNodes(node* root, vector<node*>& result, function< bool(node*) >& shouldBeErased )
{
	if(!root) return nullptr;
	
	// check if leaf node
	if( !root->left && !root->right )
	{
		if( shouldBeErased(root) )
		{
			delete root;
			return nullptr;
		}
		else
		{
			return root;
		}
	}
	
	root->left = eraseNodes( root->left, result, shouldBeErased );;
	root->right = eraseNodes(root->right, result, shouldBeErased);;
	
	if( shouldBeErased(root) )
	{
		if( root->left ) result.push_back(root->left);
		if( root->right ) result.push_back( root->right);
		return nullptr;
	}
	return root;
}

- mr.robot.fun.society November 14, 2017 | Flag Reply


Add a Comment
Name:

Writing Code? Surround your code with {{{ and }}} to preserve whitespace.

Books

is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.

Learn More

Videos

CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.

Learn More

Resume Review

Most engineers make critical mistakes on their resumes -- we can fix your resume with our custom resume review service. And, we use fellow engineers as our resume reviewers, so you can be sure that we "get" what you're saying.

Learn More

Mock Interviews

Our Mock Interviews will be conducted "in character" just like a real interview, and can focus on whatever topics you want. All our interviewers have worked for Microsoft, Google or Amazon, you know you'll get a true-to-life experience.

Learn More