This question was asked of me in an interview. How can we convert a BT such that every node in it has a value which is the sum of its child nodes?
Give each node an attached value. When you construct the tree, the value of a leaf is set; construct interior nodes to have the value leaf1.value + leaf2.value.
If you can change the values of the leaf nodes, then the operation has to go "back up" the tree updating the sum values.
This will be a lot easier if you either include back links in the nodes, or implement the tree as a "threaded tree".
Here is a solution that can help you: (the link explains it with tree-diagrams)
Convert an arbitrary Binary Tree to a tree that holds Children Sum Property
/* This function changes a tree to to hold children sum
property */
void convertTree(struct node* node)
{
int left_data = 0, right_data = 0, diff;
/* If tree is empty or it's a leaf node then
return true */
if(node == NULL ||
(node->left == NULL && node->right == NULL))
return;
else
{
/* convert left and right subtrees */
convertTree(node->left);
convertTree(node->right);
/* If left child is not present ten 0 is used
as data of left child */
if(node->left != NULL)
left_data = node->left->data;
/* If right child is not present ten 0 is used
as data of right child */
if(node->right != NULL)
right_data = node->right->data;
/* get the diff of node's data and children sum */
diff = left_data + right_data - node->data;
/* If node's data is smaller then increment node's data
by diff */
if(diff > 0)
node->data = node->data + diff;
/* THIS IS TRICKY --> If node's data is greater then increment left
subtree by diff */
if(diff < 0)
increment(node->left, -diff);
}
}
See the link to see the complete solution and explanation!
Well as Charlie pointed out, you can simply store the sum of respective subtree sizes in each inner node, and have leaves supply constant values at construction (or always implicitly use 1, if you're only interested in the number of leaves in a tree).
This is commonly known as an Augmented Search Tree.
What's interesting is that through this kind of augmentation, i.e., storing additional per-node data, you can derive other kinds of aggregate information for items in the tree as well. Any information you can express as a monoid you can store in an augmented tree, and for this, you'll need to specify:
the data type M; in your example, integers
a binary operation "op" to combine elements, with M op M -> M; in your example, the common "plus" operator
So besides subtree sizes, you can also express stuff like:
priorities (by way of the "min" or "max" operators), for efficient queries on min/max priorities;
rightmost elements in a subtree (i.e., an "op" operator that simply returns its second argument), provided that the elements you store in a tree are ordered somehow. Note that this allows us to view even regular search trees (aka. dictionaries -- "store this, retrieve that key") as augmented trees with a corresponding monoid.
(This concept is rather reminiscent of heaps, or more explicitly treaps, which store random priorities with inner nodes for probabilistic balancing. It's also quite commonly described in the context of Finger Trees, although these are not the same thing.)
If you also provide a neutral element for your monoid, you can then walk down such a monoid-augmented search tree to retrieve specific elements (e.g., "find me the 5th leaf" for your size example; "give me the leaf with the highest priority").
Uhm, anyways. Might have gotten carried away a bit there.. I just happen to find that topic quite interesting. :)
Here is the code for the sum problem. It works i have tested it.
int sum_of_left_n_right_nodes_4m_root(tree* local_tree){
int left_sum = 0;
int right_sum = 0;
if(NULL ==local_tree){
return 0;
}
if((NULL == local_tree->left)&&(NULL == local_tree->right)){
return 0;
}
sum_of_left_n_right_nodes(local_tree->left);
sum_of_left_n_right_nodes(local_tree->right);
if(NULL != local_tree->left)
left_sum = local_tree->left->data +
local_tree->left->sum;
if(NULL != local_tree->right)
right_sum = local_tree->right->data + \
local_tree->right->sum;
local_tree->sum= right_sum + left_sum;
}
With a recursive function you can do so by making the value of each node equal to the sum of the values of it's childs under condition that it has two children, or the value of it's single child if it has one child, and if it has no childs (leaf), then this is the breaking condition, the value never changes.
Related
Like the title says, I want to count the nodes in for any given level of the tree. I already know how to make member functions for counting all the nodes of the tree, just not sure how to approach a specific level. Here's what I've tried. Any help is appreciated.
First parameter is a point to a character array inputted by the user. root is a private variable representing the "oldest" node.
int TreeType::GetNodesAtLevel(ItemType* itemArray, int level)
{
TreeNode* p = root;
if (itemArray == NULL)
return;
if (level == 0)
{
cout << p->info << " ";
return;
}
else
{
GetNodesAtLevel(itemarray->left, level); //dereference in one and not the other was just testing
GetNodesAtLevel(*itemarray->right, level); //neither seems to work
}
}
The way to do it is by using a queue (employing level order traversal - BFS). Now follow this:
Take two variables, count_level and count_queue (keeping total nodes in a queue).
for a tree like this:
A
/ \
B C
/ \ \
K L D
/
E
initially count_level = 0 and count_queue = 0. Now:
Add a node to the queue(at this point A, increment count_queue to 1).
Now when you find count_level = 0 do this -> count_level = count_queue.
Add the kid nodes while dequeuing till the count_level becomes 0. So at this point the follow step 2 and that will give you the no of nodes at level beneath what just has been processed.
I'm required to write a C++ function that, given a range (a,b], returns the number of nodes in an AVL tree that are in that given range, specifically in log(n) time complexity.
I can add more fields to the tree's nodes if I need to do so.
I should point out that a,b will not necessarily appear in the tree. For example, if the tree's nodes are: 1,2,5,7,9,10, then running the function using the parameters (3,9] should return 3.
Which algorithm should I use to achieve this?
This is a famous problem - dynamic order statistcs by tree augmentation.
You basically need to augment your nodes so that when you look at a child pointer, you know how many children are in the child's subtree at time O(1). It's easy to see that this can be done without affecting the complexity.
Once you have that, you can answer any query (between this and that, inclusive/exclusive - all possibilities) by performing two traversals from node to roots. The exact traversals depend on the details (check the functions lower_bound and upper_bound in C++ for example).
First you could implement a split by key operation. That is, given a tree, to perform split(tree, key, ts, tg) splits the key in two trees; ts contains the keys less than key; t2 the greater or equal ones. This operation can be done in O(lg n).
Then, with two splits, the first on a and the second on b you can obtain the desired subset range in O(lg n).
The split could be implemented as follows (pseudo code):
void split(Node * root, const Key & key, Node *& ts, Node *& tg) noexcept
{
if (root == Node::NullPtr)
return;
if (key < KEY(root))
{
Node * r = RLINK(root), * tgaux = Node::NullPtr;
split(LLINK(root), key, ts, tgaux);
insert(tgaux, root); // insert root in tgaux
tg = join_ex(tgaux, r);
}
else
{ // ket greater or equal than key to tg
Node * l = LLINK(root), *tsaux = Node::NullPtr;
split(RLINK(root), key, tsaux, tg));
insert(tsaux, root); // insert root in tsaux
ts = join_ex(l, tsaux);
}
}
The join_ex(t1, t2) joins two exclusive trees; that is, all the keys of t1 are lesser that any key of tree t2. This join can be implemented in O(lg n) in a similar way to the concatenation described by Knuth in TAOCP V3 6.2.3.
Grosso modo if you want to join l and r, then suppose h(l) > h(r). You remove from r the leftmost node (the minimum). Let j this join node and r' the resulting tree (r - j). Now you descend by the right side of r until reaching a node p such that h(p) - h(r') equals 0 or 1. At this moment you do
And you treat p as if this was inserted.
EDIT: I was wrong in interpreting the question. Sorry. I did not see that it was to count not to calculate a set. The following would be my answer. I do not erase what I've written because I think it is useful anyway.
Ami Tavory was right.
If you use extended trees, that is to store the subtree cardinality in each node, then you could easily compute the inorder positios of a key. I usually call to this operation position(key). If key is not in the set then it returns the position that key had if it was inserted in the tree.
The inorder position of root is the cardinality of left tree.
Now, in order to count the cardinality of [a, b) set you perform position(b) - position(a). You could require to do some adjustments if a or b are not present in the tree. But basically is thus.
position(key) is, I think, "naturally" simple. Supposing that the node cardinality is accessed with COUNT(node):
long position(Node * root, const Key & key) noexcept
{
if (r == Node::NullPtr)
return 0;
if (key < KEY(root))
return position(LLINK(r), key, p);
else if (KEY(r) < key)
return position(RLINK(r), key) + COUNT(LLINK(r)) + 1;
else // the root contains key
return COUNT(LLINK(r));
}
Since an avl tree is balanced, position takes O(lg n). So two calls take O(lg n). A non recursive version is simple.
I hope you know to forgive my mistake
I was asked to implement a binary search tree with follow operation for each node v - the complexity should be O(1). The follow operation should return a node w (w > v).
I proposed to do it in O(log(n)) but they wanted O(1)
Upd. It should be next greater node
just keep the maximum element for the tree and always return it for nodes v < maximum.
You can get O(1) if you store pointers to the "next node" (using your O(log(n) algorithm), given you are allowed to do that.
How about:
int tree[N];
size_t follow(size_t v) {
// First try the right child
size_t w = v * 2 + 1;
if(w >= N) {
// Otherwise right sibling
w = v + 1;
if(w >= N) {
// Finally right parent
w = (v - 1) / 2 + 1;
}
}
return w;
}
Where tree is a complete binary tree in array form and v/w are represented as zero-based indices.
One idea is to literally just have a next pointer on each node.
You can update these pointers in O(height) after an insert or remove (O(height) is O(log n) for a self-balancing BST), which is as long as an insert or remove takes, so it doesn't add to the time complexity.
Alternatively, you can also have a previous pointer in addition to the next pointer. If you do this, you can update these pointers in O(1).
Obviously, in either case, if you have a node, you also have its next pointer, and you can simply get this value in O(1).
Pseudo-code
For only a next pointer, after the insert, you'd do:
if inserted as a right child:
newNode.next = parent.next
parent.next = newNode
else // left child
predecessor(newNode)
For both next and previous pointers:
if inserted as a right child:
parent.next.previous = newNode
newNode.next = parent.next
parent.next = newNode
else // left child
parent.previous.next = newNode
newNode.previous = parent.previous
parent.previous = newNode
(some null checks are also required).
I have a tree in which there are 3 levels. There is a root node, the root node has 3 leaf nodes and all 3 leaf nodes have 3 other leaf nodes. The nodes represent servers. Now, I have to calculate the depth of a node at for a given level. The depth is calculated as follows:
1) If a server(node) is "up" at any level and any column, then the depth of that node is 0.
2) If a server is in the last level and is "down", then depth of that node is infinity.
3) For all other cases, the depth of the node is the max depth of it's leaf nodes + 1. By max depth, it means the majority value that has occurred in it's child nodes.
A bottom up approach is followed here and hence, the depth of the root node is the depth at level 1. The level is taken as the input parameter in the program. Now, I have to calculate the depth of the root node.
I have made some assumptions regarding the program:
1) To find child nodes, follow the child pointer of the parent node.
2) To find all nodes in a given level, traverse the child nodes from root till I reach that level and make a list of them.
3) Assign the values according to the given constraints.
I am not sure whether my approach is right or not. Please help me guys. Thank you.
I think you want something along the lines of the following pseudocode:
int nodeStatus(const Node& n) {
int status = 0;
if (n.isUp)
return 1;
else if (n.isLeaf)
return -1;
else {
for (Node child : n.children)
status += nodeStatus(child);
}
if (status > 0)
return 1;
else
return -1;
}
This is a recursive method. It first checks if the node is up, in which case it returns 1. Then if the node is down and is a leaf it returns -1 as this is a failure with no children. Finally, if n is an intermediate node then it recursively calls this method again for all the children, summing the result. The final if statement then tests whether the majority of the children are classed as 1 or -1 and returns the value accordingly. Notice that by using the values 1 and -1 it's possible to just sum the children up and providing each node definitely has 3 (or an odd number of) nodes then there will never be a situation where status == 0 which would be the case where there is no majority case.
You will need to define a struct called Node somewhere that looks like this:
struct Node {
Node(bool isUp, bool isLeaf);
bool isUp;
bool isLeaf;
std::Vector<Node> children = new std::Vector<Node>(3);
}
I hope this answers your question, but it may be that I've interpreted it wrong.
I am having trouble understanding this maxDepth code. Any help would be appreciated. Here is the snippet example I followed.
int maxDepth(Node *&temp)
{
if(temp == NULL)
return 0;
else
{
int lchild = maxDepth(temp->left);
int rchild = maxDepth(temp->right);
if(lchild <= rchild)
return rchild+1;
else
return lchild+1;
}
}
Basically, what I understand is that the function recursively calls itself (for each left and right cases) until it reaches the last node. once it does, it returns 0 then it does 0+1. then the previous node is 1+1. then the next one is 2+1. if there is a bst with 3 left childs, int lchild will return 3. and the extra + 1 is the root. So my question is, where do all these +1 come from. it returns 0 at the last node but why does it return 0+1 etc. when it goes up the left/right child nodes? I don't understand why. I know it does it, but why?
Consider this part (of a bigger tree):
A
\
B
Now we want to calculate the depth of this treepart, so we pass pointer to A as its param.
Obviously pointer to A is not NULL, so the code has to:
call maxDepth for each of A's children (left and right branches). A->right is B, but A->left is obviously NULL (as A has no left branch)
compare these, choose the greatest value
return this chosen value + 1 (as A itself takes a level, doesn't it?)
Now we're going to look at how maxDepth(NULL) and maxDepth(B) are calculated.
The former is quite easy: the first check will make maxDepth return 0. If the other child were NULL too, both depths would be equal (0), and we have to return 0 + 1 for A itself.
But B is not empty; it has no branches, though, so (as we noticed) its depth is 1 (greatest of 0 for NULLs at both parts + 1 for B itself).
Now let's get back to A. maxDepth of its left branch (NULL) is 0, maxDepth of its right branch is 1. Maximum of these is 1, and we have to add 1 for A itself - so it's 2.
The point is the same steps are to be done when A is just a part of the bigger tree; the result of this calculation (2) will be used in the higher levels of maxDepth calls.
Depth is being calculated using the previous node + 1
All the ones come from this part of the code:
if(lchild <= rchild)
return rchild + 1;
else
return lchild + 1;
You add yourself +1 to the results obtained in the leaves of the tree. These ones keep adding up until you exit all the recursive calls of the function and get to the root node.
Remember in binary trees a node has at most 2 children (left and right)
It is a recursive algorithm, so it calls itself over and over.
If the temp (the node being looked at) is null, it returns 0, as this node is nothing and should not count. that is the base case.
If the node being looked at is not null, it may have children. so it gets the max depth of the left sub tree (and adds 1, for the level of the current node) and the right subtree (and adds 1 for the level of the current node). it then compares the two and returns the greater of the two.
It dives down into the two subtrees (temp->left and temp->right) and repeats the operation until it reaches nodes without children. at that point it will call maxDepth on left and right, which will be null and return 0, and then start returning back up the chain of calls.
So if you you have a chain of three nodes (say, root-left1-left2) it will get down to left2 and call maxDepth(left) and maxDepth(right). each of those return 0 (they are null). then it is back at left2. it compares, both are 0, so the greater of the two is of course 0. it returns 0+1. then we are at left1 - repeats, finds that 1 is the greater of its left n right (perhaps they are the same or it has no right child) so it returns 1+1. now we are at root, same thing, it returns 2+1 = 3, which is the depth.
Because the depth is calculated with previous node+1
To find Maximum depth in binary tree keep going left and Traveres the tree, basically perform a DFS
or
We can find the depth of the binary search tree in three different recursive ways
– using instance variables to record current depth and total depth at every level
– without using instance variables in top-bottom approach
– without using instance variables in bottom-up approach
The code snippet can be reduced to just:
int maxDepth(Node *root){
if(root){ return 1 + max( maxDepth(root->left), maxDepth(root->right)); }
return 0;
}
A good way of looking at this code is from the top down:
What would happen if the BST had no nodes? We would have root = NULL and the function would immediately return an expected depth of 0.
Now suppose the tree was populated with a number of nodes. Starting at the top, the if condition would be true for the root node. We then ask, what is the max depth of the LEFT SUB TREE and the RIGHT SUB TREE by passing the root of those sub trees to maxDepth. Both the LST and the RST of the root are one level deeper than the root, so we must add one to get the depth of the tree at root of the tree passed to the function.
i think this is the right answer
int maxDepth(Node *root){
if(root){ return 1 + max( maxDepth(root->left), maxDepth(root->right)); }
return -1;
}