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How to make an efficient algorithm for chopping a binary search tree into two by given value?

I implemented public BinarySearchTree<Node,T> chop(T x)
that chops my tree into two parts at element x. The SSet this will contain elements < x, and the returned SSet is a SSet that contains elements >= x. This should work for all elements regardless of whether they are in this.
For example, suppose s={2,4,6,8}. Then s.chop(3) returns {4,6,8} and s becomes {2}. We would get the same result for s.chop(4).
The slowChop method is implemented, but it has a time complexity of O(n), but I need to reduce it to at least O(h) when the tree is balanced (where h is the height of the tree).
public BinarySearchTree<Node,T> slowChop(T x) {
Node sample = super.newNode();
BinarySearchTree<Node,T> other = new
BinarySearchTree<Node, T>(sample);
// Iterate through the n nodes in-order.
// When see value >=x, add to new BST in O(height) time, and
// remove it from this BST (on next iteration) in O(height) time.
Iterator<T> it = iterator();
T prev = null;
while( it.hasNext() ) {
T curr = (T)(it.next());
if( c.compare(curr, x) >= 0 ) { // we have our first >= x
other.add(curr);
if( prev != null ) {
this.remove(prev); // safe to remove now
}
prev = curr;
}
}
if( prev != null ) {
this.remove(prev); // edge case, get that last one!
}
return other;
}
public BinarySearchTree<Node,T> chop(T x) {
Node sample = super.newNode();
BinarySearchTree<Node,T> other = new
BinarySearchTree<Node, T>(sample);
// TODO: Implement this method. It should match slowChop in
// behaviour, but should be faster :-)
return other;
}

Indeed, your algorithm is not making use of the efficiency that you can get from the fact you are dealing with a binary search tree. So iterating through the tree with an in-order traversal is not the way to go.
Instead, perform a binary search and cut the edges that link two nodes which should end up in different trees. While cutting you'll also need to reattach nodes to where a previous cut was performed. The complexity is the same as a binary search towards the bottom of the tree, and so it is O(logn).
Here is an implementation that assumes you have the regular getters and setters:
on the Node class: getLeft, setLeft, getRight, setRight, getValue, and
on the BinarySearchTree class: getRoot, setRoot
public BinarySearchTree<Node,T> chop(T x) {
// Create two temporary dummy (sentinel) nodes to ease the process.
Node rightRootParent = super.newNode();
Node leftRootParent = super.newNode();
// Set "cursors" at both sides
Node rightParent = rightRootParent;
Node leftParent = leftRootParent;
// Start the binary search for x, cutting edges as we walk down
Node node = this.getRoot();
while (node != null) {
// Decide for each node in the binary search path at which side it should go
if (c.compare(node.getValue(), x) >= 0) {
// Node should belong to the right-side tree
rightParent.setLeft(node); // Establish edge
rightParent = node;
node = node.getLeft(); // Move down
rightParent.setLeft(null); // Cut next edge for now (it might get restored)
} else { // Symmetric case
leftParent.setRight(node);
leftParent = node;
node = node.getRight();
leftParent.setRight(null);
}
}
// Set the roots of both trees
this.setRoot(leftRootParent.getRight());
return BinarySearchTree<Node, T>(rightRootParent.getLeft());
}

Finding the shortest path nodes with breadth first search

I am running breadth first search on the above graph to find the shortest path from Node 0 to Node 6.
My code
public List<Integer> shortestPathBFS(int startNode, int nodeToBeFound){
boolean shortestPathFound = false;
Queue<Integer> queue = new LinkedList<Integer>();
Set<Integer> visitedNodes = new HashSet<Integer>();
List<Integer> shortestPath = new ArrayList<Integer>();
queue.add(startNode);
shortestPath.add(startNode);
while (!queue.isEmpty()) {
int nextNode = queue.peek();
shortestPathFound = (nextNode == nodeToBeFound) ? true : false;
if(shortestPathFound)break;
visitedNodes.add(nextNode);
System.out.println(queue);
Integer unvisitedNode = this.getUnvisitedNode(nextNode, visitedNodes);
if (unvisitedNode != null) {
queue.add(unvisitedNode);
visitedNodes.add(unvisitedNode);
shortestPath.add(nextNode); //Adding the previous node of the visited node
shortestPathFound = (unvisitedNode == nodeToBeFound) ? true : false;
if(shortestPathFound)break;
} else {
queue.poll();
}
}
return shortestPath;
}
I need to track down the nodes through which the BFS algo. traversed to reach node 6, like [0,3,2,5,6]. For that I have created a List named shortestPath & trying to store the previous nodes of the visited nodes, to get the list of nodes. Referred
But it doesn't seem to work. The shortest path is [0,3,2,5,6]
In the list what I get is Shortest path: [0, 0, 0, 0, 1, 3, 3, 2, 5]
It's partially correct but gives the extra 1 .
If I again start from the first element 0 of the shortestPath list & start traversing & backtracking. Like 1 doesn't has an edge to 3, so I backtrack & move from 0 to 3 to 5, I will get the answer but not sure if that's the correct way.
What is the ideal way to getting the nodes for the shortest path?

Storing all the visited nodes in a single list is not helpful for finding the shortest path because in the end you have no way of knowing which nodes were the ones that led to the target node, and which ones were dead ends.
What you need to do is for every node to store the previous node in the path from the starting node.
So, create a map Map<Integer, Integer> parentNodes, and instead of this:
shortestPath.add(nextNode);
do this:
parentNodes.put(unvisitedNode, nextNode);
After you reach the target node, you can traverse that map to find the path back to the starting node:
if(shortestPathFound) {
List<Integer> shortestPath = new ArrayList<>();
Integer node = nodeToBeFound;
while(node != null) {
shortestPath.add(node)
node = parentNodes.get(node);
}
Collections.reverse(shortestPath);
}

As you can see in acheron55 answer:
"It has the extremely useful property that if all of the edges in a graph are unweighted (or the same weight) then the first time a node is visited is the shortest path to that node from the source node"
So all you have to do, is to keep track of the path through which the target has been reached.
A simple way to do it, is to push into the Queue the whole path used to reach a node, rather than the node itself.
The benefit of doing so is that when the target has been reached the queue holds the path used to reach it.
Here is a simple implementation :
/**
* unlike common bfs implementation queue does not hold a nodes, but rather collections
* of nodes. each collection represents the path through which a certain node has
* been reached, the node being the last element in that collection
*/
private Queue<List<Node>> queue;
//a collection of visited nodes
private Set<Node> visited;
public boolean bfs(Node node) {
if(node == null){ return false; }
queue = new LinkedList<>(); //initialize queue
visited = new HashSet<>(); //initialize visited log
//a collection to hold the path through which a node has been reached
//the node it self is the last element in that collection
List<Node> pathToNode = new ArrayList<>();
pathToNode.add(node);
queue.add(pathToNode);
while (! queue.isEmpty()) {
pathToNode = queue.poll();
//get node (last element) from queue
node = pathToNode.get(pathToNode.size()-1);
if(isSolved(node)) {
//print path
System.out.println(pathToNode);
return true;
}
//loop over neighbors
for(Node nextNode : getNeighbors(node)){
if(! isVisited(nextNode)) {
//create a new collection representing the path to nextNode
List<Node> pathToNextNode = new ArrayList<>(pathToNode);
pathToNextNode.add(nextNode);
queue.add(pathToNextNode); //add collection to the queue
}
}
}
return false;
}
private List<Node> getNeighbors(Node node) {/* TODO implement*/ return null;}
private boolean isSolved(Node node) {/* TODO implement*/ return false;}
private boolean isVisited(Node node) {
if(visited.contains(node)) { return true;}
visited.add(node);
return false;
}
This is also applicable to cyclic graphs, where a node can have more than one parent.

In addition to the already given answer by user3290797.
It looks like You are dealing with an unweighted graph. We interpret this as every edge has a weight of 1. In this case, once You have associated a distance to the root node with every node of the graph (the breadth-first traversal), it becomes trivial to reconstruct the shortest path from any node, and even detect if there are multiple ones.
All You need to do is a breadth- (in case You want every shortest path) or depth-first traversal of the same graph starting from the target node and only considering neighbours with a depth's value of exactly 1 less.
So we need to jump from distance 4 (node 6) to 3, 2, 1, 0, and there is only one way (in this case) to do so.
In case we are interested in the shortest path to node 4 the result would be distances 2-1-0 or nodes 4-3-0 or 4-8-0.
BTW, this approach can easily be modified to work with weighted graphs (with non-negative weights) too: valid neighbours are those with distance equals to current minus the weight of the edge -- this involves some actual calculations and directly storing previous nodes along the shortest path might be better.

Depth first search list paths to all end nodes

Hi I have a tree in which I would like to get paths from the initial (root) node to all leaves.
I found several algortithms that list (all) apths betwenn any given two nodes within a graph (for example this SO question:
Graph Algorithm To Find All Connections Between Two Arbitrary Vertices)
For binary tree there also exists an algorithm
http://techieme.in/print-all-paths-in-a-tree/
but I work on a tree with various branching factors.
Is there any better way of achieving what I want to do than traversing the tree once in order to get all leaves and then run the algorithm above for all leaves combined with the initial node?
I was thinking about implementing simple DFS extended by some additional stack containing all nodes alongt he path to a single leaf and then listing all sentences by looping through these stacks.
ArrayList<GrammarNode> discovered = new ArrayList<GrammarNode>();
Stack<GrammarNode> S = new Stack<GrammarNode>();
while (!S.empty()) {
node = S.pop();
if (!discovered.contains(node)) {
discovered.add(node);
System.out.println(node.getWord.getSpelling().trim());
for (GrammarArc arc : node.getSuccessors()) {
S.push(arc.getGrammarNode());
}
}
}
UPDATE:
The problem of this is that one has alyways go back to the root in order to generate full sentences. So I guess the question is: How to remember the node which was already fully visited (this means where all children nodes were already explored)?

Printing all paths from the root to every leaf would mean to print the entire tree so I'd just use a simple DFS and do the following for each node:
add it to the list/stack
if the node has children, repeat for the children
if the node is a leaf, print the list/stack
pop the node from the list/stack
Example:
A
/ \
B E
/ \ / \
C D F G
The first steps would look like this:
put A on the list -> {A}
put B on the list -> {A,B}
put C on the list -> {A,B,C}
since C is a leaf, print the list (A,B,C)
remove C from the list -> {A,B}
put D on the list -> {A,B,D}
since D is a leaf, print the list (A,B,D)
...

if you know that the graph is indeed a tree (there is only one path to each node), them yes, a simple DFS would be more efficient (at least from a memory usage point of view). Otherwise, you can also use the iterative deepening DFS.

So here's a sample approach. Note that you need an extra visited field in your node structure:
public class TreeNodeExtra {
int val;
TreeNodeExtra left;
TreeNodeExtra right;
boolean visited;
TreeNodeExtra (int v) {
val = v;
visited = false;
}
}
private ArrayList<ArrayList<TreeNodeExtra>> all_path_from_root_to_leaf(TreeNodeExtra root) {
Stack<TreeNodeExtra> st = new Stack<>();
ArrayList<ArrayList<TreeNodeExtra>> res = new ArrayList<>();
st.push(root);
root.visited = true;
while (!st.isEmpty()) {
TreeNodeExtra top = st.peek();
if (top.left != null && !top.left.visited) {
st.push(top.left);
top.left.visited = true;
}
// if left node is null
else {
if (top.right == null && top.left == null) {
// we have a leaf
ArrayList<TreeNodeExtra> tmpList = new ArrayList<>();
for (TreeNodeExtra t : st) {
tmpList.add(t);
}
res.add(tmpList);
st.pop();
}
else if (top.right != null && !top.right.visited) {
st.push(top.right);
top.right.visited = true;
}
else {
st.pop();
}
}
}
return res;
}

A slight modification of DFS (which includes back-tracking) prints all the paths from a given source. In the below example the graph is represented in adjacency list format.
public void mDFS(ArrayList<node> v,int ind,ArrayList<Boolean> visit,ArrayList<node> tillNow){
visit.set(ind,true);
node n = v.get(ind);
int len = n.adj.size();
tillNow.add(n);
int count = 0;
for(node tmp: n.adj){
if( !visit.get(tmp.id) ){
count++;
tmp.pre = ind;
mDFS(v,tmp.id,visit,tillNow); // id gives index of node in v
}
}
if(count == 0){
for(node tmp: tillNow){
System.out.print((tmp.id + 1) + " - ");
}System.out.print("\n");
}
visit.set(ind,false);
tillNow.remove(tillNow.size() - 1);
return;
}

I need to delete a node in a linked list, how do I find the node that comes BEFORE it so I can rearrange the list?

It's for a school assignment. I have to use a search method that returns the node that I search for or the one right before it if it doesn't exist. Obviously if I want to delete a node it'll return that one node and I won't be able to find the one that comes before it. Here's the code for the search method:
private myNode search(myEntry searchEntry)
{
myNode ref = first;
myNode pre = null;
while(ref != null)
{
if(searchEntry.compareTo(ref.data) < 0)
break;
pre = ref;
ref = ref.link;
}
return pre;
}
first is the first node, ref is the pointer, pre is the node preceding the pointer.
Maybe I'll use a doubly-linked list if it doesn't require me to rewrite things too much but if there's a simple way to find the predecessor of the node I'm trying to delete using this search method then I'd like to know. I'm not supposed to be using doubly-linked lists at all.

You can't do that with your search method. You have to implement a distinct search method that return the predecessor of the node and then delete it.

If you need to delete ref, than you return pre. To delete ref, you can write something like
pre = list.search(entry); //find the prescending node
ref = pre.link; //get the node you want to delete
pre.link = ref.link; //reassign link
ref.delete(); //delete node

Shortest path (fewest nodes) for unweighted graph

I'm trying build a method which returns the shortest path from one node to another in an unweighted graph. I considered the use of Dijkstra's but this seems a bit overkill since I only want one pair. Instead I have implemented a breadth-first search, but the trouble is that my returning list contains some of the nodes that I don't want - how can I modify my code to achieve my goal?
public List<Node> getDirections(Node start, Node finish){
List<Node> directions = new LinkedList<Node>();
Queue<Node> q = new LinkedList<Node>();
Node current = start;
q.add(current);
while(!q.isEmpty()){
current = q.remove();
directions.add(current);
if (current.equals(finish)){
break;
}else{
for(Node node : current.getOutNodes()){
if(!q.contains(node)){
q.add(node);
}
}
}
}
if (!current.equals(finish)){
System.out.println("can't reach destination");
}
return directions;
}

Actually your code will not finish in cyclic graphs, consider graph 1 -> 2 -> 1. You must have some array where you can flag which node's you've visited already. And also for each node you can save previous nodes, from which you came. So here is correct code:
private Map<Node, Boolean>> vis = new HashMap<Node, Boolean>();
private Map<Node, Node> prev = new HashMap<Node, Node>();
public List getDirections(Node start, Node finish){
List directions = new LinkedList();
Queue q = new LinkedList();
Node current = start;
q.add(current);
vis.put(current, true);
while(!q.isEmpty()){
current = q.remove();
if (current.equals(finish)){
break;
}else{
for(Node node : current.getOutNodes()){
if(!vis.contains(node)){
q.add(node);
vis.put(node, true);
prev.put(node, current);
}
}
}
}
if (!current.equals(finish)){
System.out.println("can't reach destination");
}
for(Node node = finish; node != null; node = prev.get(node)) {
directions.add(node);
}
directions.reverse();
return directions;
}

Thank you Giolekva!
I rewrote it, refactoring some:
The collection of visited nodes doesn't have to be a map.
For path reconstruction, the next node could be looked up, instead of the previous node, eliminating the need for reversing the directions.
public List<Node> getDirections(Node sourceNode, Node destinationNode) {
//Initialization.
Map<Node, Node> nextNodeMap = new HashMap<Node, Node>();
Node currentNode = sourceNode;
//Queue
Queue<Node> queue = new LinkedList<Node>();
queue.add(currentNode);
/*
* The set of visited nodes doesn't have to be a Map, and, since order
* is not important, an ordered collection is not needed. HashSet is
* fast for add and lookup, if configured properly.
*/
Set<Node> visitedNodes = new HashSet<Node>();
visitedNodes.add(currentNode);
//Search.
while (!queue.isEmpty()) {
currentNode = queue.remove();
if (currentNode.equals(destinationNode)) {
break;
} else {
for (Node nextNode : getChildNodes(currentNode)) {
if (!visitedNodes.contains(nextNode)) {
queue.add(nextNode);
visitedNodes.add(nextNode);
//Look up of next node instead of previous.
nextNodeMap.put(currentNode, nextNode);
}
}
}
}
//If all nodes are explored and the destination node hasn't been found.
if (!currentNode.equals(destinationNode)) {
throw new RuntimeException("No feasible path.");
}
//Reconstruct path. No need to reverse.
List<Node> directions = new LinkedList<Node>();
for (Node node = sourceNode; node != null; node = nextNodeMap.get(node)) {
directions.add(node);
}
return directions;
}

You must include the parent node to each node when you put them on your queue. Then you can just recursively read the path from that list.
Say you want to find the shortest path from A to D in this Graph:
/B------C------D
/ |
A /
\ /
\E---------
Each time you enqueue a node, keep track of the way you got here.
So in step 1 B(A) E(A) is put on the queue. In step two B gets dequeued and C(B) is put on the queue etc. Its then easy to find your way back again, by just recursing "backwards".
Best way is probably to make an array as long as there are nodes and keep the links there, (which is whats usually done in ie. Dijkstra's).

Every time through your loop, you call
directions.Add(current);
Instead, you should move that to a place where you really know you want that entry.

It is really no simpler to get the answer for just one pair than for all the pairs. The usual way to calculate a shortest path is to start like you do, but make a note whenever you encounter a new node and record the previous node on the path. Then, when you reach the target node, you can follow the backlinks to the source and get the path. So, remove the directions.add(current) from the loop, and add code something like the following
Map<Node,Node> backlinks = new HashMap<Node,Node>();
in the beginning and then in the loop
if (!backlinks.containsKey(node)) {
backlinks.add(node, current);
q.add(node);
}
and then in the end, just construct the directions list in backwards using the backlinks map.