I want to insert in the tree but not using any other data structures like queue. I want to insert in level order and no matter what I code, it doesn't. Also I couldn't find any code without queues or things like that.
Here is my attempt;
void insert(int x) {
if (root == NULL) {
root = new node(x, NULL, NULL);
return;
}
node *temp = root;
node *prev = root;
while (temp != NULL) {
if (temp->left != NULL) {
prev = temp;
temp = temp->left;
} else if (temp->right != NULL) {
prev = temp;
temp = temp->right;
}
}
if (temp->left == NULL)
prev->left = new node(x, NULL, NULL);
else if (temp->right == NULL)
prev->right = new node(x, NULL, NULL);
}
I don't have a link for recursive insertion but it should work like this:
bool recursion(node * current_node, node * to_insert, int &max_depth, int cur_depth) {
if(max_depth < cur_depth) {
max_depth = cur_depth;
}
for (auto & current_child : {current_node->left, current_node->right})
if(current_child == NULL) {
if( max_depth > cur_depth ) {
current_child -> left = to_insert;
return true;
}
} else {
if(recursion(current_child, to_insert, max_depth, cur_depth + 1)) {
return true;
}
}
return false;
}
This does depth-first-search (not breadth-first, I was mistaken above, they are very similar in trees) from left to right. So we will first find the left-most leaf, then the one right next to it and so on. We will always track how deep we are in the tree. If at one point we find a node on the second deepest layer that hasn't got a child, it will add the node we want to insert at this point and recurse up the tree. Due to the order in which we traverse the tree, this will find the left most open spot, so exactly what you want.
This method can return false if the submost layer of the tree is full. Then we have to go down to the left-most leaf and insert the node at its left child. One can also save this leaf somehow when we first find it, but that seemed more complicate to me then just searching it again (this can be done without problem in a for-loop).
You can replace the recursive method by an iteration with a stack (there are many sources on the internet explaining how to make a recursive depth-first-search to a iterative one).
I don't really like the in-out-parameter max_depth but it was the easiest to do this.
Related
Would someone mind briefly explaining why I keep getting segmentation faults from this code?
"target" is the Node to be deleted, and I'm trying to replace it with the rightmost leaf on the left branch. "parent" is the parent of target.
else if (target->left != NULL && target->right != NULL) {
Node * childishDad = target;
Node * childish = target->left;
while(childish->right != NULL){
childishDad = childish;
childish = childish->right;
}
if(parent->left == target)
parent->left = childish;
if(parent->right == target)
parent->right = childish;
childish->left = target->left;
childish->right = target->right;
if(childishDad != target)
childishDad->right = NULL;
delete target;
return true;
}
When the while-loop completes, you know
childish->right == NULL
However, later you assign
childish->right = target->right;
Maybe that's the problem.
I realized that I wasn't checking whether parent == NULL, so it was failing in the case where I tried to delete the root node.
Helly everyone,
I took a C++ coding course with practically no prior knowledge(my understanding of pointers is still somewhat shakey)at University this semester.
I have to implement a binary search tree in C++ and my problem is as follows:
Over a predefined Node structure with values and pointers to a left and a right node I am supposed to implement several functions, two of them being:
void InsertNode(Node* root, Node* node)
which is supposed to fit the handed node into the given tree "root",
and another one called
void InsertValue(Node* root, int value)
which should create a new instance of the node struct with the passed value and fit it in the given tree "root". To do so I am supposed to use both CreateNode (simple function to create Node* pointer of a new Node instance with int value and left/right pointers set to NULL) and InsertNode.
Im kind of running in a treadmill here and i dont think i really understand how the functions are supposed to work(eg. the difference between them).
Yesterday i wrote this function:
void InsertNode(Node* root, Node* node){
if(root == NULL){
root = CreateNode(node->value);
}
else if(root->value < node->value){
if(node->left != NULL){
InsertNode(root, node->left);
}
else{
node->left = CreateNode(root->value);
}
}
else if(root->value > node->value){
if(node->right != NULL){
InsertNode(root, node->right);
}
else{
node->right = CreateNode(root->value);
}
}
}
Since im not really able to test these functions without the later functions that will actually build the tree with given nodes i was curious if i could get some help here with the next functions InsertValue(what is it supposed to do that InsertNode doesnt do already? :S)
Greetings and thanks in advance.
Initial note: This answer assumes that the InsertNode function is initially called with root being the root of the tree, and node being the node to insert into the tree.
One problem is this statement:
root = CreateNode(node->value);
Since the argument root is passed by value, which means that it is copied, the assignment will only change the local copy. Once the function returns the original pointer that you pass to the function will not have changed.
You need to pass the pointer by reference, meaning the root argument references the original variable passed in to the function, instead of it being copied. You do this by using an ampersand when declaring the argument:
Node*& root
The above means that root is a reference to a pointer to Node.
So the complete InsertNode declaration should look like
void InsertNode(Node*& root, Node* node)
There are also other problems, for example these lines are not correct:
if(node->left != NULL){
InsertNode(root, node->left);
}
else{
node->left = CreateNode(root->value);
}
This is not correct because node->left should be NULL always, which makes you create a new node using the value from the root of the tree, and assign it to node->left, but you never insert node in the tree.
What you should instead do is simply
InsertNode(node->left, node);
Of course you should do the same change for setting the right branch.
Combining the two solutions above, your function would look like
void InsertNode(Node*& root, Node* node)
{
if (root == 0)
root = node;
else if (root->value < node->value)
InsertNode(root->left, node);
else
InsertNode(root->right, node);
}
This function also solves a third problem with your current code: What if node->value is equal to root->value? The above function puts it in the right branch.
When you are creating a tree, value are also assigned with each node. See following code:
typedef struct BST {
int data;
struct BST *lchild, *rchild;
} node;
void insert(node *root, node *new_node) {
if (new_node->data < root->data) {
if (root->lchild == NULL)
root->lchild = new_node;
else
insert(root->lchild, new_node);
}
if (new_node->data > root->data) {
if (root->rchild == NULL)
root->rchild = new_node;
else
insert(root->rchild, new_node);
}
}
node *new_node, *root;
int main()
{
new_node = get_node();
printf("\nEnter The Element ");
scanf("%d", &new_node->data);
if (root == NULL) /* Tree is not Created */
root = new_node;
else
insert(root, new_node)
}
The below code is in Python and is used for insertion in a BST ::
class Node :
def __init__(self.key):
self.left = None
self.right = None
self.val = key
def insert(root.node):
if root is None :
root = node
else :
if root.val < node.val:
if root.right is None :
root.right = node
else :
insert(root.right, node)
else :
if root.left is None :
root.left = node
else :
insert(root.left, node)
def inorder(root):
if root :
inorder(root.left)
print(root.val)
inorder(root.right)
# Driver program to test the above functions
# Let us create the following BST
# 50
# / \
# 30 70
# / \ / \
# 20 40 60 80
r = Node(50)
insert(r,Node(30))
insert(r,Node(20))
insert(r,Node(40))
insert(r,Node(70))
insert(r,Node(60))
insert(r,Node(80))
# Print inoder traversal of the BST
inorder(r)
I have this function for deleting a node in a binary search tree which seems to be working EXCEPT in the case where I ask it to delete the root node. It is supposed to take the right-most value on the left and replace the node with that; however, once that happens, the new root node's children pointers don't seem to point to the original root node's children. Code is as follows:
bool delete_node(Node*& root, TYPE data) {
Node* toDelete;
Node* parent;
// This function is defined appropriately elsewhere, and finds the target to be deleted
toDelete = find(data, root);
if (!toDelete) {
return false;
}
// This function is defined appropriately elsewhere, and finds the parent of the node to be deleted
parent = find_parent(root, toDelete);
// Other cases left out because they work
// If the target node has two children:
if (toDelete->left && toDelete->right)
{
// find rightmost child on left that is a leaf
Node *replacement = toDelete->left;
while (replacement->right)
{
replacement = replacement->right;
}
// set the target node's data
toDelete->data = replacement->data;
if (parent)
{
if ( parent->data < toDelete->data )
{
parent->right = replacement;
} else
{
parent->left = replacement;
}
} else
{
// if node has no parents, then it is the root and should be replaced with replacement
// This line here is what seems to be causing my trouble...I think
root = replacement;
}
parent = find_parent(toDelete, replacement);
if (parent)
{
if (parent->left == replacement)
parent->left = NULL;
else
parent->right = NULL;
}
delete toDelete;
return true;
}
}
Thanks in advance!
what I ended coming up with was this: keep track of the parent node that is one above the node that replaces the node to be deleted. there will then be 2 cases to consider: the parent is the node to be deleted and parent is not the node to be deleted. by replacing the appropriate parts of the tree at the right case, the structure and invariants of the tree remained ok and the node to be deleted was successfully deleted. technically, it would be the data at the node to be deleted.
else if (toDelete->left != NULL && toDelete->right != NULL) {
// find rightmost child on left that is a leaf
Node* replacement = toDelete->left;
parent = toDelete;
// parent is now the parent of the replacement
while ( replacement->right ) {
parent = replacement;
replacement = replacement->right;
} // By the end, parent will be the node one above replacement
toDelete->key = replacement->key;
if (parent == target)
parent->left = replacement->left;
else
parent->right = replacement->left;
delete replacement;
return true;
}
This is what I did to make it work. Just check if the node is the root node, and if so, set the new root. Below is the working code I have. The three places marked by asterisks is what I added to make it work. All the other lines of code is just standard textbook theory.
inline NamesBinaryTree::Node* NamesBinaryTree::Node::removeNode (Node*& node, const Female* female, stringComparisonFunction s) { // Taken from p.253 of Alex Allain's "Jumping Into C++".
if (!node)
return nullptr;
if (node->femaleInfo.first == female->getName()) {
if (!node->left) { // i.e. node has either one child or zero children.
Node* rightNode = node->right;
if (node->isRoot()) // ***
namesBinaryTree.setRoot(rightNode); // Tested to work correctly. Note that we cannot call 'delete node;', since that will delete the very root that we are setting!
else
delete node;
return rightNode; // This will return nullptr if node->right is also nullptr, which is what we would want to do anyway since that would mean that node has zero children.
}
if (!node->right) { // i.e. node has exactly one child, namely its left child, in which case return that left child.
Node* leftNode = node->left;
if (node->isRoot()) // ***
namesBinaryTree.setRoot(leftNode);
else
delete node;
return leftNode; // This will never be nullptr, else the previous if condition would have been met instead.
}
Node* maxNode = findMaxNode(node->left); // node has two children, so it shall be replaced by the largest valued node in its left subtree.
maxNode->left = removeMaxNode(node->left, maxNode); // Note that maxNode->left = node->left is not enough because without actually removing maxNode, the node that was pointing to maxNode will now be pointing to maxNode in its new position (and the subtree under it), and the subtree that was under maxNode will now be gone.
maxNode->right = node->right;
if (node->isRoot()) // ***
namesBinaryTree.setRoot(maxNode); // Tested to work correctly.
else
delete node;
return maxNode;
}
else {
const int result = (*s)(female->getName(), node->femaleInfo.first);
if (result < 0)
node->left = removeNode(node->left, female, s); // This assignment can only work if node is passed by reference (hence the parameter Node*& node), at least this is what "C++ Essentials" did in their solution, p.247.
else // Don't use 'else if (result > 0)'. Let the equality case be covered here too (just as in NamesBinaryTree::Node::insertNode).
node->right = removeNode(node->right, female, s); // Again, this assignment can only work if node is passed by reference (hence the parameter Node*& node).
}
return node; // So if node->femaleInfo.first != female->getName(), then the same node is returned, which means that the two assignment lines above don't change any values.
}
Edited*: I'm working on the delete function for a binary search tree. I'm just working on the first case right now. I think this is correct, but I'm wondering if it can be done recursively, or more efficiently. Any help is appreciated. Assume BSTSearch searches for a node, isLeaf returns true if the node is a leaf, and each node has a pointer that allows them access to their parent.
void
BinarySearchTree::BSTDelete(int x, BSTNode *node){
BSTNode *deleteNode;
deleteNode = BSTSearch(x,node);
if(isLeaf(deleteNode)){
if(deleteNode->sortkey > (deleteNode->parent)->sortkey){
delete (deleteNode->parent)->right;
(deleteNode->parent)->right = NULL;
}
else{
delete (deleteNode->parent)->left;
(deleteNode->parent)->left = NULL;
}
}
You don't need a pointer to the parent. Here is a recursive version that should work: (pass by reference (&), in case you don't know, allows you to modify the variable, similar to pass by pointer; BSTNode *& is a pointer passed by reference, so we can modify the value of node->left/right (pointers) (not just what they're pointing to))
void BinarySearchTree::BSTDelete(int x, BSTNode *&node)
{
if (node == NULL)
return;
if (x == node->sortKey)
{
if (isLeaf(node))
{
delete node;
node = NULL;
}
else
{
// other stuff goes here
}
return;
}
else if (x < node->sortKey)
BSTDelete(x, node->left);
else
BSTDelete(x, node->right);
}
I am writing a program that adds, deletes, and displays nodes (that are doubly linked) and their components, but whenever I try to retrieve a node and display it's components I get this error:
2 [main] a 4640 exception::handle: Exception: STATUS_ACCESS_VIOLATION
2875 [main] a 4640 open_stackdumpfile: Dumping stack trace to a.exe.stackdump
I have narrowed it down to the search function within my .h file that is supposed to search to see if there is a node within the linked list that account number being searched. The function returns the node that comes before it, or the "previous" node.
Here is my search function:
bool searchListByAcctNum (int searchKey, nodePtr *prevOut)
{
bool found = false;
nodePtr p = headNum;
nodePtr prev = NULL;
while (p != NULL)
{
if (p->acctNum < searchKey)
{
prev = p;
p = p->nextNum;
}
else
{
if (p->acctNum == searchKey)
found = true;
p = NULL;
}
}
*prevOut = prev;
return found;
If anyone could help me at all, I'd appreciate it!
It looks like your list may be corrupted, or the pointer you're passing to receive the previous node is invalid, since that code looks okay. However, it seems to me that it could be written in a much simpler manner:
bool searchListByAcctNum (int searchKey, nodePtr *prevOut) {
/// Start at beginning of list, use pointer variable to hold previous.
nodePtr p = headNum;
*prevOut = = NULL;
// Process entire list, will exit early if need be.
while (p != NULL) {
// If past it, just return false, caller should ignore prevOut.
if (p->acctNum > searchKey)
return false;
// If equal, return true, prevOut holds previous or NULL if found at start.
if (p->acctNum == searchKey) {
return true;
// Save previous and advance to next.
*prevOut = p;
p = p->next;
}
// Reached end of list without finding, caller should ignore prevOut.
return false;
}