I am trying to preorder a BST
I am not sure how to do it.
You should consider a recursive approach rather than an iterative one. Tree traversal (preorder, inorder, and postorder) is very easily done using recursion.
The Wikipedia article on tree traversal has a pseudocode recursive algorithm, but there's really not much to it. Since you're storing the tree in an array, you won't have node pointers, just indices.
As for how you know when you reach the leaf nodes, well, their indices will be beyond the end of the array.
Doing a tree traversal is usually done with recursive functions because that is the most natural way to express the traversal.
Doing a pre-order traversal is pretty easy. Ignoring the details, you have something like this:
void do_preorder(Tree t)
{
// do something with the current tree node
if (leftnode(t) not empty)
do_preorder(leftnode(t));
if (rightnode(t) not empty)
do_preorder(rightnode(t))
}
If you want to be tricky, you can even create a generic traversal that allows you to choose the flavor (pre-order, in-order or post-order) at runtime:
void do_xorder(Tree t, Flavor f)
{
if (f == PREORDER)
handle_currentnode(t)
if (leftnode(t) not empty)
do_xorder(leftnode(t), f);
if (f == INORDER)
handle_currentnode(t)
if (rightnode(t) not empty)
do_xorder(rightnode(t), f)
if (f == POSTORDER)
handle_currentnode(t)
}
As per you code you can use (2i + 1) for left and (2i + 2) for right child.
My suggestion that you should start storing elements from index 1 in the array because you can store left child at ( 2 * i) index and right child at (2*i + 1) index .
And end of the left child can be easily found by checking whether ( 2*i > currSize ) is true.
http://en.wikipedia.org/wiki/Tree_traversal
Check out this link for any traversal you want to perform on tree...
Related
I use the following method to traverse* a binary tree of 300 000 levels:
Node* find(int v){
if(value==v)
return this;
else if(right && value<v)
return right->find(v);
else if(left && value>v)
return left->find(v);
}
However I get a segmentation fault due to stack overflow.
Any ideas on how to traverse the deep tree without the overhead of recursive function calls?
*
By "traverse" I mean "search for a node with given value", not full tree traversal.
Yes! For a 300 000 level tree avoid recursion. Traverse your tree and find the value iteratively using a loop.
Binary Search Tree representation
25 // Level 1
20 36 // Level 2
10 22 30 40 // Level 3
.. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. // Level n
Just to clarify the problem further. Your tree has a depth of n = 300.000 levels. Thus, in the worst case scenario a Binary Search Tree (BST) will have to visit ALL of the tree's nodes. This is bad news because that worst case has an algorithmic O(n) time complexity. Such a tree can have:
2ˆ300.000 nodes = 9.9701e+90308 nodes (approximately).
9.9701e+90308 nodes is an exponentially massive number of nodes to visit. With these numbers it becomes so clear why the call stack overflows.
Solution (iterative way):
I'm assuming your Node class/struct declaration is a classic standard integer BST one. Then you could adapt it and it will work:
struct Node {
int data;
Node* right;
Node* left;
};
Node* find(int v) {
Node* temp = root; // temp Node* value copy to not mess up tree structure by changing the root
while (temp != nullptr) {
if (temp->data == v) {
return temp;
}
if (v > temp->data) {
temp = temp->right;
}
else {
temp = temp->left;
}
}
return nullptr;
}
Taking this iterative approach avoids recursion, hence saving you the hassle of having to recursively find the value in a tree so large with your program call stack.
A simple loop where you have a variable of type Node* which you set to the next node, then loop again ...
Don't forget the case that the value you are searching for does not exist!
You could implement the recursion by not using the call stack but a user-defined stack or something similar; this could be done via the existing stack template. The approach would be to have a while loop which iterates until the stack is empty; as the existing implementaion uses depth-first search, elimination of the recursive calls can be found here.
When the tree that you have is a Binary Search Tree, and all you want to do is search for a node in it that has a specific value, then things are simple: no recursion is necessary, you can do it using a simple loop as others have pointed out.
In the more general case of having a tree which is not necessarily a Binary Search Tree, and wanting to perform a full traversal of it, the simplest way is using recursion, but as you already understand, if the tree is very deep, then recursion will not work.
So, in order to avoid recursion, you have to implement a stack on the C++ heap. You need to declare a new StackElement class that will contain one member for each local variable that your original recursive function had, and one member for each parameter that your original recursive function accepted. (You might be able to get away with fewer member variables, you can worry about that after you have gotten your code to work.)
You can store instances of StackElement in a stack collection, or you can simply have each one of them contain a pointer to its parent, thus fully implementing the stack by yourself.
So, instead of your function recursively calling itself, it will simply consist of a loop. Your function enters the loop with the current StackElement being initialized with information about the root node of your tree. Its parent pointer will be null, which is another way of saying that the stack will be empty.
In every place where the recursive version of your function was calling itself, your new function will be allocating a new instance of StackElement, initializing it, and repeating the loop using this new instance as the current element.
In every place where the recursive version of your function was returning, your new function will be releasing the current StackElement, popping the one that was sitting on the top of the stack, making it the new current element, and repeating the loop.
When you find the node you were looking for, you simply break from the loop.
Alternatively, if the node of your existing tree supports a) a link to its "parent" node and b) user data (where you can store a "visited" flag) then you don't need to implement your own stack, you can just traverse the tree in-place: in each iteration of your loop you first check if the current node is the node you were looking for; if not, then you enumerate through children until you find one which has not been visited yet, and then you visit it; when you reach a leaf, or a node whose children have all been visited, then you back-track by following the link to the parent. Also, if you have the freedom to destroy the tree as you are traversing it, then you do not even need the concept of "user data": once you are done with a child node, you free it and make it null.
Well, it can be made tail recursive at the cost of a single additional local variable and a few comparisons:
Node* find(int v){
if(value==v)
return this;
else if(!right && value<v)
return NULL;
else if(!left && value>v)
return NULL;
else {
Node *tmp = NULL;
if(value<v)
tmp = right;
else if(value>v)
tmp = left;
return tmp->find(v);
}
}
Walking through a binary tree is a recursive process, where you'll keep walking until you find that the node you're at currently points nowhere.
It is that you need an appropriate base condition. Something which looks like:
if (treeNode == NULL)
return NULL;
In general, traversing a tree is accomplished this way (in C):
void traverse(treeNode *pTree){
if (pTree==0)
return;
printf("%d\n",pTree->nodeData);
traverse(pTree->leftChild);
traverse(pTree->rightChild);
}
So I am having trouble finding the correct ancestor in a 2-3 Tree. In a 2-3 Tree of arbitrary height, there are a couple of cases to look for.
My nodes are designed as follows:
template<typename DataType>
struct node{
Node<DataType> *child1; //left-child
Node<DataType> *child2; //middle-child (3-node only)
Node<DataType> *child3; //right-child
Node<DataType> *extraChild; //for splitting purposes
Node<DataType> *parent;
DataType key1; //key1 <= key2
DataType key2; //key2 <= extraKey (only when node needs to be split)
DataType extraKey; //for splitting purposes
};
Is there an algorithm or something similar to find the proper ancestor to a node?
For example, say we were looking for the ancestor of H (bottom of the provided tree), it is obvious from a visual perspective that ancestor of H is the H at the root of the tree. But that requires jumping up 4 parent links. The tree could be of arbitrary size which is the issue.
My goal in the end is to create an iterator that does an in order traversal of the 2-3 tree. The purpose of finding an ancestor node is that the ancestor node will be the in-order successor of the leaf node that is the right child of it's parent node. Again, as in the example provided above.
If the goal is simply to traverse a 2-3 tree in-order, then you just need to write a recursive function that starts at the root.
The algorithm is as follows:
traverse(Node n)
if n.left != null
traverse(n.left)
if n.middle != null
visit(n.key1)
traverse(n.middle)
visit(n.key2)
else
visit(n.key1)
if n.right != null
traverse(n.right)
Algorithm taken from this link.
I have a balanced binary search tree of integers and I want to find the leftmost node which stores the integer greater or equal to a fixed number like a using a function like ask(a).
for example suppose that I have added the following points in my tree, 8,10,3,6,1,4,7,14,13
Then the tree would be like this:
now ask(1) should be 1, ask(3) should be 3, ask(2) should be 3 and so on.
I think that I can use Inorder traversal to write my ask function, But I don't know how.
Iv written this piece of code so far:
inorderFind(node->left, a);
if (node->key.getX() >= a)
return node;
inorderFind(node->right, a);
The first argument is the current tree node and a is the a that is described above. I know that I can use a bool variable like flag and set it to true when the if condition holds, and then it would prevent from walking through other nodes of the tree and returning a false node. Is there anything else that I can do?
Trees have the wonderful property of allowing queries through simple, recursive algorithms. So, let's try to find a recursive formulation of your query.
Say LEFTMOST(u) is a function which answers this question :
Given the binary search subtree rooted at node u, with(possibly null) left and
right children l and r, respectively, what is the left-most node
with a value >= a?
The relation is quite simple:
LEFTMOST(u) = LEFTMOST(l) if it exists
LEFTMOST(r) otherwise
That's it. How you translate this to your problem and how you handle concepts like "null" and "does not exist" is a function of your representation.
I have implement a link-based BST (binary search tree) in C++ for one of my assignment. I have written my whole class and everything works good, but my assignment asks me to plot the run-times for:
a. A sorted list of 50000, 75000, and 100000 items
b. A random list of 50000, 75000, and 100000 items
That's fine, I can insert the numbers but it also asks me to call the FindHeight() and CountLeaves() methods on the tree. My problem is that I've implemented the two functions using recursion. Since I have a such a big list of numbers I'm getting getting a stackoverflow exception.
Here's my class definition:
template <class TItem>
class BinarySearchTree
{
public:
struct BinarySearchTreeNode
{
public:
TItem Data;
BinarySearchTreeNode* LeftChild;
BinarySearchTreeNode* RightChild;
};
BinarySearchTreeNode* RootNode;
BinarySearchTree();
~BinarySearchTree();
void InsertItem(TItem);
void PrintTree();
void PrintTree(BinarySearchTreeNode*);
void DeleteTree();
void DeleteTree(BinarySearchTreeNode*&);
int CountLeaves();
int CountLeaves(BinarySearchTreeNode*);
int FindHeight();
int FindHeight(BinarySearchTreeNode*);
int SingleParents();
int SingleParents(BinarySearchTreeNode*);
TItem FindMin();
TItem FindMin(BinarySearchTreeNode*);
TItem FindMax();
TItem FindMax(BinarySearchTreeNode*);
};
FindHeight() Implementation
template <class TItem>
int BinarySearchTree<TItem>::FindHeight()
{
return FindHeight(RootNode);
}
template <class TItem>
int BinarySearchTree<TItem>::FindHeight(BinarySearchTreeNode* Node)
{
if(Node == NULL)
return 0;
return 1 + max(FindHeight(Node->LeftChild), FindHeight(Node->RightChild));
}
CountLeaves() implementation
template <class TItem>
int BinarySearchTree<TItem>::CountLeaves()
{
return CountLeaves(RootNode);
}
template <class TItem>
int BinarySearchTree<TItem>::CountLeaves(BinarySearchTreeNode* Node)
{
if(Node == NULL)
return 0;
else if(Node->LeftChild == NULL && Node->RightChild == NULL)
return 1;
else
return CountLeaves(Node->LeftChild) + CountLeaves(Node->RightChild);
}
I tried to think of how I can implement the two methods without recursion but I'm completely stumped. Anyone have any ideas?
Recursion on a tree with 100,000 nodes should not be a problem if it is balanced. The depth would only be maybe 17, which would not use very much stack in the implementations shown. (log2(100,000) = 16.61). So it seems that maybe the code that is building the tree is not balancing it correctly.
I found this page very enlightening because it talks about the mechanics of converting a function that uses recursion to one that uses iteration.
It has examples showing code as well.
May be you need to calculate this while doing the insert. Store the heights of nodes, i.e add an integer field like height in the Node object. Also have counters height and leaves for the tree. When you insert a node, if its parent is (was) a leaf, the leaf count doesnt change, but if not, increase leaf count by 1. Also the height of the new node is parent's height + 1, hence if that is greater than the current height of the tree, then update it. Its a homework, so i wont help with the actual code
Balance your tree occasionally. If your tree is getting stackoverflow on FindHeight(), that means your tree is way unbalanced. If the tree is balanced it should only have a depth of about 20 nodes for 100000 elements.
The easiest (but fairly slow) way of re-balancing unbalanced binary tree is to allocate an array of TItem big enough to hold all of the data in the tree, insert all of your data into it in sorted order, and delete all of the nodes. Then rebuild the tree from the array recursively. The root is the node in the middle. root->left is the middle of the left half, root->right is the middle of the right half. Repeat recursively. This is the easiest way to rebalance, but it is slowish and takes lots of memory temporarily. On the other hand, you only have to do this when you detect that the tree is very unbalanced, (depth on insert is more than 100).
The other (better) option is to balance during inserts. The most intuitive way to do this is to keep track of how many nodes are beneath the current node. If the right child has more than twice as many "child" nodes as the left child, "rotate" left. And vice-versa. There's instrcutions on how to do tree rotates all over the internet. This makes inserts slightly slower, but then you don't have occassional massive stalls that the first option creates. On the other hand, you have to constantly update all of the "children" counts as you do the rotates, which isn't trivial.
In order to count the leaves without recursion, use the concept of an iterator like the STL uses for the RB-tree underlying std::set and std::map ... Create a begin() and end() function for you tree that indentifies the ordered first and last node (in this case the left-most node and then the right-most node). Then create a function called
BinarySearchTreeNode* increment(const BinarySearchTreeNode* current_node)
that for a given current_node, will return a pointer to the next node in the tree. Keep in mind for this implementation to work, you will need an extra parent pointer in your node type to aid in the iteration process.
Your algorithm for increment() would look something like the following:
Check to see if there is a right-child to the current node.
If there is a right-child, use a while-loop to find the left-most node of that right subtree. This will be the "next" node. Otherwise go to step #3.
If there is no right-child on the current node, then check to see if the current node is the left-child of its parent node.
If step #3 is true, then the "next" node is the parent node, so you can stop at this point, otherwise go the next step.
If the step #3 was false, then the current node is the right-child of the parent. Thus you will need to keep moving up to the next parent node using a while loop until you come across a node that is a left-child of its parent node. The parent of this left-child node will then be the "next" node, and you can stop.
Finally, if step #5 returns you to the root, then the current node is the last node in the tree, and the iterator has reached the end of the tree.
Finally you'll need a bool leaf(const BinarySearchTreeNode* current_node) function that will test whether a given node is a leaf node. Thus you counter function can simply iterate though the tree and find all the leaf nodes, returning a final count once it's done.
If you want to measure the maximum depth of an unbalanced tree without recursion, you will, in your tree's insert() function, need to keep track of the depth that a node was inserted at. This can simply be a variable in your node type that is set when the node is inserted in the tree. You can then iterate through the three, and find the maximum depth of a leaf-node.
BTW, the complexity of this method is unfortunately going to be O(N) ... nowhere near as nice as O(log N).
In this question I'm not asking how to do it but HOW IS IT DONE.
I'm trying (as an excersise) implement simple map and although I do not have problems with implementing links and they behavior (how to find next place to insert new link etc.) I'm stuck with the problem how to implement iterating over a map. When you think about it and look at std::map this map is able to return begin and end iterator. How? Especially end?
If map is a tree how can you say which branch of this map is an end? I just do not understand it. An how to iterate over a map? Starting from the top of the tree and then what? Go and list everything on the left? But those nodes on the left have also links to the right. I really don't know. I will be really glad if someone could explain it to me or give me a link so I could read about it.
A map is implemented using a binary search tree. To meet the complexity requirements it has to be a self-balancing tree, so a red-black tree is usually used, but that doesn't affect how you iterate over the tree.
To read the elements out of a binary search tree in order from least to greatest, you need to perform an in-order traversal of the tree. The recursive implementation is quite simple but isn't really practical for use in an iterator (the iterator would have to maintain a stack internally, which would make it relatively expensive to copy).
You can implement an iterative in-order traversal. This is an implementation taken from a library of tree containers I wrote a while ago. NodePointerT is a pointer to a node, where the node has left_, right_, and parent_ pointers of type NodePointerT.
// Gets the next node in an in-order traversal of the tree; returns null
// when the in-order traversal has ended
template <typename NodePointerT>
NodePointerT next_inorder_node(NodePointerT n)
{
if (!n) { return n; }
// If the node has a right child, we traverse the link to that child
// then traverse as far to the left as we can:
if (n->right_)
{
n = n->right_;
while (n->left_) { n = n->left_; }
}
// If the node is the left node of its parent, the next node is its
// parent node:
else if (n->parent_ && n == n->parent_->left_)
{
n = n->parent_;
}
// Otherwise, this node is the furthest right in its subtree; we
// traverse up through its parents until we find a parent that was a
// left child of a node. The next node is that node's parent. If
// we have reached the end, this will set node to null:
else
{
while (n->parent_ && n == n->parent_->right_) { n = n->parent_; }
n = n->parent_;
}
return n;
}
To find the first node for the begin iterator, you need to find the leftmost node in the tree. Starting at the root node, follow the left child pointer until you encounter a node that has no left child: this is the first node.
For an end iterator, you can set the node pointer to point to the root node or to the last node in the tree and then keep a flag in the iterator indicating that it is an end iterator (is_end_ or something like that).
The representation of your map's iterator is totally up to you. I think it should suffice to use a single wrapped pointer to a node. E.g.:
template <typename T>
struct mymapiterator
{
typename mymap<T>::node * n;
};
Or something similar. Now, mymap::begin() could return such instance of the iterator that n would point to the leftmost node. mymap::end() could return instance with n pointing to root probably or some other special node from which it is still possible to get back to rightmost node so that it could satisfy bidirectional iteration from end iterator.
The operation of moving between the nodes (operators++() and operator--(), etc.) are about traversing the tree from smaller to bigger values or vice versa. Operation that you probably have already implemented during insertion operation implementation.
For sorting purposes, a map behaves like a sorted key/value container (a.k.a. a dictionary); you can think of it as a sorted collection of key/value pairs, and this is exactly what you get when you query for an iterator. Observe:
map<string, int> my_map;
my_map["Hello"] = 1;
my_map["world"] = 2;
for (map<string, int>::const_iterator i = my_map.begin(); i != my_map.end(); ++i)
cout << i->first << ": " << i->second << endl;
Just like any other iterator type, the map iterator behaves like a pointer to a collection element, and for map, this is a std::pair, where first maps to the key and second maps to the value.
std::map uses a binary search internally when you call its find() method or use operator[], but you shouldn't ever need to access the tree representation directly.
One big trick you may be missing is that the end() iterator does not need to point to anything. It can be NULL or any other special value.
The ++ operator sets an iterator to the same special value when it goes past the end of the map. Then everything works.
To implement ++ you might need to keep next/prev pointers in each node, or you could walk back up the tree to find the next node by comparing the node you just left to the parent's right-most node to see if you need to walk to that parent's node, etc.
Don't forget that the iterators to a map should stay valid during insert/erase operations (as long as you didn't erase the iterator's current node).