As we all know, when inserting into a complete binary tree we have to fill all the children for all the leafs from left to right. I have the following method that inserts a node into a complete binary tree.
//fields
private T item;
private int size;
private CBTree<T> left, right;
//add method
public void add(T item)
{
if(left == null)
{
left = new CBTree<T>(item);
size += left.size;
}
else if(right == null)
{
right = new CBTree<T>(item);
size += right.size;
}
else if(!((left.left != null) && (left.right != null)) &&
((right.left == null) || (right.right == null)))
{
left.add(item);
}
else
{
right.add(item);
}
}
The problem with this implementation is that after the 11th node it adds the 12th node to the left child of 8 instead of 6. I understand that this is happening because the following line is reassigning the root of this tree to be the left child of the root.
left.add(item);
So it is changing the root to 2. Is there a way to change the root back to its original value? I am trying to accomplish this without using stacks and queues.
It's not sufficient to just check children of children to determine which side to go to, because as soon as the tree reaches height 4 that won't work any more, since children of children of the root won't change from that point forward yet we can still go either left or right.
2 approaches come to mind:
Have a complete variable at each node.
A node with no children is complete.
A node with 2 complete children of equal size is complete.
Whenever update the tree (insert or delete) you update this variable for each affected node as well.
Mathematically determine whether a subtree is complete based on the size.
A tree of size 2^n - 1 is complete (for some n).
Note: this will only work if we're not allowed to freely delete elements without keeping the tree complete.
For either approach, when doing an insertion, we go left (left.add(item)) if either of these conditions are true:
the left subtree is not complete
the left and right subtrees are of the same size (both complete, meaning we're increasing the height with this insertion)
I'll leave the implementation details to you.
Note: you need to also update size when doing left.add(item); and right.add(item);. You could probably just stick a size++ in the add function, since we're adding 1 element so size increases by 1 no matter what.
Thanks to Dukeling's answer, the correct way to implement the method was to mathematically determine if the subtree was full. Here is the code:
//fields
private T item;
private int size;
private CBTree<T> left, right;
//add method
public void add(T item)
{
if(left == null)
{
left = new CBTree<T>(item);
}
else if(right == null)
{
right = new CBTree<T>(item);
}
else if(leftFull())
{
right.add(item);
}
else
{
left.add(item);
}
size++;
}
//Checks if the left subtree is full
public boolean leftFull()
{
int used, leafs = 1;
while(leafs <= size + 1)
{
leafs *= 2;
}
leafs /= 2;
used = (size + 1) % leafs;
if(used >= (leafs / 2))
{
return true;
}
else
{
return false;
}
}
Related
I want to implement dfs for nodes that are of type long in Java.
My algorithm calculates correctly the number of nodes, and the number
of edges, but not the sequence of nodes. Could you please help me
modify my algorithm so I calculate the order in which the nodes are
visited, correctly?
This is my code:
private int getNumberOfNodes(long firstNode) {
List<Long> marked = new ArrayList<>(); //------------------------------------------->
Stack<Long> stack = new Stack<Long>(); //step 1 Create/declare stack
stack.push(firstNode); //Step 2 Put/push inside the first node
while (!stack.isEmpty()) { //Repeat till stack is empty:
Long node = stack.pop(); //Step 3 Extract the top node in the stack
marked.add(node); //------------------------------------------->
long[] neighbors = xgraph.getNeighborsOf(node); //Get neighbors
if (neighbors.length % 2 == 0) {
} else {
numOfNodesWithOddDegree++;
}
int mnt = 0;
for (long currentNode : neighbors) {
if (!marked.contains(currentNode) && !stack.contains(currentNode) ) { //&& !stack.contains(currentNode)
stack.push(currentNode);
} else {
}
if (!marked.contains(currentNode)) {
numOfEdges++;
}
}
}
return marked.size(); //(int) Arrays.stream(neighbors).count();
}
I guess you exam the marked list for the sequence.
As your graph is undirected, the sequence of traversals could be varied based on which neighbor you pushed into the stack first. which means the logic of your function:
xgraph.getNeighborsOf(node)
could impact your sequence. see Vertex orderings from this wiki https://en.wikipedia.org/wiki/Depth-first_search
so my conclusion is: you may have a different traversal sequence, it does not mean your DFS is wrong, as long as it is Deep first search, it is ok to be a little bit different from the given answer.
What's the difference between these two versions?
public static int countLeaves(IntTreeNode root) {
if (root == null) {
return 0;
} else
return 1 + countLeaves(root.left) + countLeaves(root.right);
}
public static int countLeaves(IntTreeNode root) {
if (root == null) {
return 0;
} else if (root.left == null && root.right == null) {
return 1;
} else
return countLeaves(root.left) + countLeaves(root.right);
}
I couldn't find anything that use the first version in the internet.
Is the first version wrong?
I have tried to trace them on paper, they seem to be the same.
But I just want to be sure.
The first seems to count all nodes in a tree, whereas the second one one counts all leafs.
Indeed in the first one, the recursion stops when there is no valid tree anymore (root == null) and it always goes into recursion checking the left and right tree by adding 1 (for the current node).
The second only counts the leafs using the condition if (root.left == null && root.right == null).
That's assuming a leaf is identified as node that has a null root.left and a null root.right.
The first version is not counting leaves - it's counting nodes.
The second version is indeed counting leaves.
These methods will not return the same result, here's an example:
root(5)
/ \
leaf(3) leaf(7)
for such a tree the first method will return 3 (number of nodes) and the second one will return 2 (number of leafs).
I'm trying to answer the following programming question:
In the heap.java program, the insert() method inserts a new node in the heap and ensures the heap condition is preserved. Write a toss() method that places a new node in the heap array without attempting to maintain the heap condition. (Perhaps each new item can simply be placed at the end of the array.) Then write a restoreHeap() method that restores the heap condition throughout the entire heap. Using toss() repeatedly followed by a single restoreHeap() is more efficient than using insert() repeatedly when a large amount of data must be inserted at one time. See the description of heapsort for clues. To test your program, insert a few items, toss in some more, and then restore the heap.
I've written the code for the toss function which successfully inserts the node at the end and doesn't modify the heap condition. I'm having problems with the restoreHeap function though and I can't wrap my head around it. I've included the two functions below.
The full code of heap.java is here (includes toss() and restoreHeap() )
toss() - I based this off the insert function
public boolean toss(int key)
{
if(currentSize==maxSize)
return false;
Node newNode = new Node(key);
heapArray[currentSize] = newNode;
currentSize++;
return true;
} // end toss()
restoreHeap() - I based this off the trickleUp function and I'm getting a StackOverflowError.
public void restoreHeap(int index)
{
int parent = (index-1) / 2;
Node bottom = heapArray[index];
while( index > 0 &&
heapArray[parent].getKey() < bottom.getKey() )
{
heapArray[index] = heapArray[parent]; // move it down
index = parent;
parent = (parent-1) / 2;
} // end while
heapArray[index] = bottom;
while(index != 0)
{
restoreHeap(parent++);
}
} // end restoreHeap()
Any ideas? Help appreciated.
I'll give it a shot. Here is a way to do what you asked with some explanation.
Since you know that half of all nodes in a heap are leafs and a leaf, by itself, is a valid heap, you only have to run through the other half of the nodes to make sure they also are valid. If we do this from the bottom and up, we can maintain a valid heap structure "below" as we go up through the heap. This can easily be accomplished by a for loop:
public void rebuildHeap()
{
int half = heapArray.length / 2;
for(int i = half; i >= 0; i--)
restoreHeap(i);
}
How is restoreHeap implemented then?
It's supposed to check the node at index against its children to see if it needs to relocate the node. Because we make sure that the trees below the index node are heaps, we only have to move the index node to the right position. Hence we move it down in the tree.
First we need to locate the children. Since each row in the three have twice as many nodes as the row before, the children can be located like this:
private void restoreHeap(int index)
{
int leftChild = (index * 2) + 1; //+1 because arrays start at 0
int rightChild = leftChild +1;
...
Now you just have to compare the childrens value against your index nodes value. If a child have a bigger value you need to swap the index node with the child node. If both children have a bigger value, you need to swap with the child with the biggest value of the two (to maintain the heap structure after the swap). When the nodes have been swapped you need to call the method again to see if you need to move the index node further down the tree.
...
int biggest = index;
if(leftChild < currentSize && heapArray[leftChild].getKey() > heapArray[index].getKey())
biggest = leftChild; //LeftChild is bigger
if(rightChild < currentSize && heapArray[rightChild].getKey() > heapArray[biggest].getKey())
biggest = rightChild; //RightChild is bigger than both leftChild and the index node
if(biggest != index) //If a swap is needed
{
//Swap
Node swapper = heapArray[biggest];
heapArray[biggest] = heapArray[index];
heapArray[index] = swapper;
restoreHeap(biggest);
}
}
I have a n-ary tree which contains key values (integers) in each node. I would like to calculate the minimum depth of the tree. Here is what I have come up with so far:
int min = 0;
private int getMinDepth(Node node, int counter, int temp){
if(node == null){
//if it is the first branch record min
//otherwise compare min to this value
//and record the minimum value
if(counter == 0){
temp = min;
}else{
temp = Math.min(temp, min);
min = 0;
}
counter++;//counter should increment by 1 when at end of branch
return temp;
}
min++;
getMinDepth(node.q1, counter, min);
getMinDepth(node.q2, counter, min);
getMinDepth(node.q3, counter, min);
getMinDepth(node.q4, counter, min);
return temp;
}
The code is called like so:
int minDepth = getMinDepth(root, 0, 0);
The idea is that if the tree is traversing down the first branch (branch number is tracked by counter), then we set the temp holder to store this branch depth. From then on, compare the next branch length and if it smaller, then make temp = that length. For some reason counter is not incrementing at all and always staying at zero. Anyone know what I am doing wrong?
I think you're better off doing a breadth-first search. Your current implementation tries to be depth-first, which means it could end up exploring the whole tree if the branches happen to be in an awkward order.
To do a breadth-first search, you need a queue (a ArrayDeque is probably the right choice). You'll then need a little class that holds a node and a depth. The algorithm goes a little something like this:
Queue<NodeWithDepth> q = new ArrayDeque<NodeWithDepth>();
q.add(new NodeWithDepth(root, 1));
while (true) {
NodeWithDepth nwd = q.remove();
if (hasNoChildren(nwd.node())) return nwd.depth();
if (nwd.node().q1 != null) q.add(new NodeWithDepth(nwd.node().q1, nwd.depth() + 1));
if (nwd.node().q2 != null) q.add(new NodeWithDepth(nwd.node().q2, nwd.depth() + 1));
if (nwd.node().q3 != null) q.add(new NodeWithDepth(nwd.node().q3, nwd.depth() + 1));
if (nwd.node().q4 != null) q.add(new NodeWithDepth(nwd.node().q4, nwd.depth() + 1));
}
This looks like it uses more memory than a depth-first search, but when you consider that stack frames consume memory, and that this will explore less of the tree than a depth-first search, you'll see that's not the case. Probably.
Anyway, see how you get on with it.
You are passing the counter variable by value, not by reference. Thus, any changes made to it are local to the current stack frame and are lost as soon as the function returns and that frame is popped of the stack. Java doesn't support passing primitives (or anything really) by reference, so you'd either have to pass it as a single element array or wrap it in an object to get the behavior you're looking for.
Here's a simpler (untested) version that avoids the need to pass a variable by reference:
private int getMinDepth(QuadTreeNode node){
if(node == null)
return 0;
return 1 + Math.min(
Math.min(getMinDepth(node.q1), getMinDepth(node.q2)),
Math.min(getMinDepth(node.q3), getMinDepth(node.q4)));
}
Both your version and the one above are inefficient because they search the entire tree, when really you only need to search down to the shallowest depth. To do it efficiently, use a queue to do a breadth-first search like Tom recommended. Note however, that the trade-off required to get this extra speed is the extra memory used by the queue.
Edit:
I decided to go ahead and write a breadth first search version that doesn't assume you have a class that keeps track of the nodes' depths (like Tom's NodeWithDepth). Once again, I haven't tested it or even compiled it... But I think it should be enough to get you going even if it doesn't work right out of the box. This version should perform faster on large, complex trees, but also uses more memory to store the queue.
private int getMinDepth(QuadTreeNode node){
// Handle the empty tree case
if(node == null)
return 0;
// Perform a breadth first search for the shallowest null child
// while keeping track of how deep into the tree we are.
LinkedList<QuadTreeNode> queue = new LinkedList<QuadTreeNode>();
queue.addLast(node);
int currentCountTilNextDepth = 1;
int nextCountTilNextDepth = 0;
int depth = 1;
while(!queue.isEmpty()){
// Check if we're transitioning to the next depth
if(currentCountTilNextDepth <= 0){
currentCountTilNextDepth = nextCountTilNextDepth;
nextCountTilNextDepth = 0;
depth++;
}
// If this node has a null child, we're done
QuadTreeNode node = queue.removeFirst();
if(node.q1 == null || node.q2 == null || node.q3 == null || node.q4 == null)
break;
// If it didn't have a null child, add all the children to the queue
queue.addLast(node.q1);
queue.addLast(node.q2);
queue.addLast(node.q3);
queue.addLast(node.q4);
// Housekeeping to keep track of when we need to increment our depth
nextCountTilNextDepth += 4;
currentCountTilNextDepth--;
}
// Return the depth of the shallowest node that had a null child
return depth;
}
Counter is always staying at zero because primitives in java are called by value. This means if you overwrite the value in a function call the caller won't see the change. Or if you're familiar with C++ notation it's foo(int x) instead of foo(int& x).
One solution would be to use an Integer object since objects are call-by-reference.
Since you're interested in the minimum depth a breadth first solution will work just fine, but you may get memory problems for large trees.
If you assume that the tree may become rather large an IDS solution would be the best. This way you'll get the time complexity of the breadth first variant with the space complexity of a depth first solution.
Here's a small example since IDS isn't as well known as its brethren (though much more useful for serious stuff!). I assume that every node has a list with children for simplicity (and since it's more general).
public static<T> int getMinDepth(Node<T> root) {
int depth = 0;
while (!getMinDepth(root, depth)) depth++;
return depth;
}
private static<T> boolean getMinDepth(Node<T> node, int depth) {
if (depth == 0)
return node.children.isEmpty();
for (Node<T> child : node.children)
if (getMinDepth(child, depth - 1)) return true;
return false;
}
For a short explanation see http://en.wikipedia.org/wiki/Iterative_deepening_depth-first_search
Recently I have been asked one question that in a singularly linked list how do we go to the middle of the list in one iteration.
A --> B --> C --> D (even nodes)
for this it should return address which points to B
A --> B --> C (odd nodes)
for this also it should return address which points to B
There is one solution of taking two pointers one moves one time and other moves two times but it does not seem working here
LinkedList p1,p2;
while(p2.next != null)
{
p1 = p1.next;
p2 = p2.next.next;
}
System.out.print("middle of the node" + p1.data); //This does not give accurate result in odd and even
Please help if anyone has did this before.
The basic algorithm would be
0 Take two pointers
1 Make both pointing to first node
2 Increment first with two node and first if its successful then traverse second to one node ahead
3 when second reaches end first one would be at middle.
Update:
It will definitely work in odd case, for even case you need to check one more condition if first point is allowed to move next but not next to next then both pointers will be at middle you need to decide which to take as middle
You can't advance p1 unless you successfully advanced p2 twice; otherwise, with a list length of 2 you end up with both pointing at the end (and you indicated even length lists should round toward the beginning).
So:
while ( p2.next != null ) {
p2 = p2.next;
if (p2.next != null) {
p2 = p2.next;
p1 = p1.next;
}
}
I know you've already accepted an answer, but this whole question sounds like an exercise in cleverness rather than an attempt to get the correct solution. Why would you do something in O(n) when you can do it in O(n/2)?
EDIT: This used to assert O(1) performance, and that is simply not correct. Thanks to ysth for pointing that out.
In practice, you would do this in zero iterations:
LinkedList list = ...
int size = list.size();
int middle = (size / 2) + (size % 2 == 0 ? 0 : 1) - 1; //index of middle item
Object o = list.get(middle); //or ListIterator it = list.listIterator(middle);
The solution of taking two pointers and one moves a half the rate should work fine. Most likely it is not the solution but your actual implementation that is the problem. Post more details of your implementation.
static ListElement findMiddle(ListElement head){
ListElement slower = head;
ListElement faster = head;
while(faster.next != null && faster.next.next !=null ){
faster = faster.next.next;
slower = slower.next;
}
return slower;
}
public static Node middle(Node head){
Node slow=head,fast=head;
while(fast!=null& fast.next!=null && fast.next.next!=null){
slow=slow.next;
fast=fast.next.next;
}
if(fast!=null && fast.next!=null){
slow=slow.next;
}
return slow;
}
public ListNode middleNode(ListNode head) {
if(head == null) return head;
ListNode slow = head;
ListNode fast = head;
while(fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
}
return slow;
}