I have a C problem which i need to check if a given Tree is a Binary Search Tree or not and then find which node is incorrect and delete it (from the problem definition, it will always be just one wrong node). For instance:
10
|
|
9 12
| |
| |
7 3 11 14
In this tree, element 3 is incorrect and should be deleted. In case of given tree is already a BST, algorithm should do nothing. I'm trying to adapt a solution from CodeForGeeks which verifies if a tree is a BST or not to instead returning a boolean, returning a pointer for the node which is wrong, but this isn't working well. Here's the method i came up with:
NODE* verifyBinaryTree(NODE* root, NO** previous){
if(node != NULL){
NODE* left= verifyBinaryTree(root->left, previous);
if(left != NULL){
return left;
}
if(*previous && root->data <= (*previous)->data) return root;
*previous = root;
return(verifyBinaryTree(root->right, previous));
}
return NULL;
}
I would enjoy if someone could help me to find the right logic behind this problem.
Here is an easy way to approach this problem . If you do inorder traversal on a valid BST , you will get the nodes sorted in increasing order .
You can do this inorder traversal and after storing them serially in a data structure (perhaps array) , check if the adjacent elements are not in increasing order . If you find such elements , you need to delete them .
I am having trouble understanding how this bit of code sorts a linked list.
node* sort(node *head) {
struct node* point;
struct node* small;
struct node* stay;
int temp;
stay = head;
while (stay != NULL) {
point = stay->next;
small = stay;
while (point != NULL) {
if (point->data < small->data) {
small = point;
}
point = point->next;
}
temp = stay->data;
stay->data = small->data;
small->data = temp;
stay = stay->next;
}
return head;
}
I have tried to follow it along on paper and my thought process leads me to believe that if we were to run this function, a list would be sorted like this:
5 -> 2 -> 1 -> 3
2 -> 5 -> 1 -> 3
2 -> 1 -> 5 -> 3
2 -> 1 -> 3 -> 5
My understanding is that the first while loop traverses the list each time until it reaches the last node, while the second while loop compares the two nodes point and small. If the data needs to be switched, the next block of code actually does the switching, and then stay moves on to the next node in the list, with point being the node after that. How does the code know to go back to the very first node and keep comparing, so that 2 gets switched with 1? Thank you for your help.
This piece of code implements selection sort: Starting from stay (small == stay), it searches for the least value following and as soon as found (i. e. end of list reached) swaps.
Be aware that in case of stay being smallest, it is swapped with itself (you could prevent this with an appropriate test before: if(small != stay) { /* swap */ }.
So actually, your sorting steps are as follows:5 -> 2 -> 1 -> 3
1 -> 2 -> 5 -> 3
1 -> 2 -> 5 -> 3 (second node swapped with itself)
1 -> 2 -> 3 -> 5
1 -> 2 -> 3 -> 5 (fourth node swapped with itself)
Actually, there is one step more, as the last node always is swapped with itself (while(stay != NULL) stops only after last node).
First node is treated correctly right from the start (in the first run of the outer loop) as stay is initially set to head.
Tried exploring a lot over the net, but could get any help,
Everywhere its like adding a node to the Binary Search tree.
Question: Requesting for algorithm and code snippet for adding a node to the Binary tree. ( or point me to correct URL )
Assumption:
As per my understanding, Binary Tree and Binary Search Tree is different? Correct me if I am wrong.
( request: if you are writing your code snippet please use proper variable name, that helps in understanding )
Eg: Binary Tree
5 7 3 x1 x2 x3
5
7 3
x1 x2 x3
Binary Search Tree 5 7 3 2 4 6
5
3 7
2 4 6
insert(int key, struct node **root)
{
if( NULL == *root )`
{
*root = (struct node*) malloc( sizeof( struct node ) );`
(*root)->data = key;
(*root)->left = NULL;
(*root)->right = NULL;
}
else if(key < (*root)->data)
{
insert( key, &(*root)->left );
}
else if(key > (*root)->data)
{
insert( key, &(*root)->right );
}
}
The difference between a Binary Tree and a Binary Search Tree is that though they both have restrictions that each node can have at most 2 child nodes, a Binary Search Tree (BST) also must have its left child be of equal or lesser value and the its right child must be of greater or equal value. This is why it is called a "Search" tree because everything is ordered numerically and it has an O(logn) run time for searching.
Because there isn't the requirement of being a BST, a Binary Tree can be stored in a vector (array). As you insert into the vector you build the Binary Tree in level-order fashion. The code is below:
// typedef the node struct to NODE
// nodeVector similar to STL's vector class
insert(int key, NODE** nodeVector)
{
NODE *newNode = (NODE*) malloc( sizeof( NODE ) );
newNode->data = key;
newNode->left = NULL;
newNode->right = NULL;
// add newNode to end of vector
int size = nodeVector->size();
nodeVector->push_back(newNode);
// if newNode is not root node
if(nodeVector->size() > 1)
{
// set parent's child values
Node* parent = (size/2)-1; // take advantage of integer division instead of using floor()
if (parent->left == NULL)
{
parent->left = newNode;
}
else
{
parent->right = newNode;
}
}
}
A Queue data structure can be used for inserting element in to a Binary Tree, since in Binary Tree the order of nodes is not maintained so we will insert the node as soon as we find any null.
Using Queue we will be traversing the Binary Tree in Level Order Traversal.
struct Treenode* temp;
Q = CreateQueue();
EnQueue(Q,root);
while(!IsEmptyQueue(Q))
{
temp = DeQueue(Q);
if(temp->left)
EnQueue(Q,temp->left);
else
{
temp->left=newNode;
DeleteQueue(Q);
return;
}
if(temp->right)
EnQueue(Q,temp->right);
else
{
temp->right=newNode;
DeleteQueue(Q);
return;
}
}
Since, I cannot comment I am writing this.
The above answer for binary tree insert function is wrong.
Suppose for 0, 1, 2 , 3, 4 , 5 passed in sequence to insert function,
its generating tree like
0
/
1
\
2
/
3
\
4
/
5`<br/>
of which inorder traversal will be 1 3 5 4 2 0
while answer should be
0
/ \
1 2
/ \ /
3 4 5
of which inorder traversal will be 3 1 4 0 5 2.
Since I also face the same problem, I came up with following solution over the net :-
You can use a queue to store the current node where we want to place the new node as we do in level order traversal and then we insert the nodes level by level.
Following link might help you :-
http://www.geeksforgeeks.org/linked-complete-binary-tree-its-creation/
I am posting this as answer because I dont have the necessary reputation to post a comment. Except for bagelboy all others have misunderstood the tree as either Binary Search Tree or Complete Binary Tree. The question is simple Binary Tree and Bagelboy's answer looks correct.
How can I detect that whether a singly linked-list has loop or not??
If it has loop then how to find the point of origination of the loop i.e. the node from which the loop has started.
You can detect it by simply running two pointers through the list, this process is known as the tortoise and hare algorithm after the fable of the same name:
First off, check if the list is empty (head is null). If so, no cycle exists, so stop now.
Otherwise, start the first pointer tortoise on the first node head, and the second pointer hare on the second node head.next.
Then loop continuously until hare is null (which may be already true in a one-element list), advancing tortoise by one and hare by two in each iteration. The hare is guaranteed to reach the end first (if there is an end) since it started ahead and runs faster.
If there is no end (i.e., if there is a cycle), they will eventually point to the same node and you can stop, knowing you have found a node somewhere within the cycle.
Consider the following loop which starts at 3:
head -> 1 -> 2 -> 3 -> 4 -> 5
^ |
| V
8 <- 7 <- 6
Starting tortoise at 1 and hare at 2, they take on the following values:
(tortoise,hare) = (1,2) (2,4) (3,6) (4,8) (5,4) (6,6)
Because they become equal at (6,6), and since hare should always be beyond tortoise in a non-looping list, it means you've discovered a cycle.
The pseudo-code will go something like this:
def hasLoop (head):
return false if head = null # Empty list has no loop.
tortoise = head # tortoise initially first element.
hare = tortoise.next # Set hare to second element.
while hare != null: # Go until hare reaches end.
return false if hare.next = null # Check enough left for hare move.
hare = hare.next.next # Move hare forward two.
tortoise = tortoise.next # Move tortoise forward one.
return true if hare = tortoise # Same means loop found.
endwhile
return false # Loop exit means no loop.
enddef
The time complexity for this algorithm is O(n) since the number of nodes visited (by tortoise and hare) is proportional to the number of nodes.
Once you know a node within the loop, there's also an O(n) guaranteed method to find the start of the loop.
Let's return to the original position after you've found an element somewhere in the loop but you're not sure where the start of the loop is.
head -> 1 -> 2 -> 3 -> 4 -> 5
^ |
| V
8 <- 7 <- 6
\
x (where hare and tortoise met).
This is the process to follow:
Advance hare and set size to 1.
Then, as long as hare and tortoise are different, continue to advance hare, increasing size each time. This eventually gives the size of the cycle, six in this case.
At this point, if size is 1, that means you must already be at the start of the cycle (in a cycle of size one, there is only one possible node that can be in the cycle so it must be the first one). In this case, you simply return hare as the start, and skip the rest of the steps below.
Otherwise, set both hare and tortoise to the first element of the list and advance hare exactly size times (to the 7 in this case). This gives two pointers that are different by exactly the size of the cycle.
Then, as long as hare and tortoise are different, advance them both together (with the hare running at a more sedate pace, the same speed as the tortoise - I guess it's tired from its first run). Since they will remain exactly size elements apart from each other at all times, tortoise will reach the start of the cycle at exactly the same time as hare returns to the start of the cycle.
You can see that with the following walkthrough:
size tortoise hare comment
---- -------- ---- -------
6 1 1 initial state
7 advance hare by six
2 8 1/7 different, so advance both together
3 3 2/8 different, so advance both together
3/3 same, so exit loop
Hence 3 is the start point of the cycle and, since both those operations (the cycle detection and cycle start discovery) are O(n) and performed sequentially, the whole thing taken together is also O(n).
If you want a more formal proof that this works, you can examine the following resources:
a question on our sister site;
the Wikipedia cycle detection page; or
"The Tortoise and the Hare Algorithm" by Peter Gammie, April 17, 2016.
If you're simply after support for the method (not formal proof), you can run the following Python 3 program which evaluates its workability for a large number of sizes (how many elements in the cycle) and lead-ins (elements before the cycle start).
You'll find it always finds a point where the two pointers meet:
def nextp(p, ld, sz):
if p == ld + sz:
return ld
return p + 1
for size in range(1,1001):
for lead in range(1001):
p1 = 0
p2 = 0
while True:
p1 = nextp(p1, lead, size)
p2 = nextp(nextp(p2, lead, size), lead, size)
if p1 == p2:
print("sz = %d, ld = %d, found = %d" % (size, lead, p1))
break
The selected answer gives an O(n*n) solution to find the start node of the cycle. Here's an O(n) solution:
Once we find the slow A and fast B meet in the cycle, make one of them still and the other continue to go one step each time, to decide the perimeter of the cycle, say, P.
Then we put a node at the head and let it go P steps, and put another node at the head. We advance these two nodes both one step each time, when they first meet, it's the start point of the cycle.
You can use hash map also to finding whether a link list have loop or not below function uses hash map to find out whether link list have loop or not
static bool isListHaveALoopUsingHashMap(Link *headLink) {
map<Link*, int> tempMap;
Link * temp;
temp = headLink;
while (temp->next != NULL) {
if (tempMap.find(temp) == tempMap.end()) {
tempMap[temp] = 1;
} else {
return 0;
}
temp = temp->next;
}
return 1;
}
Two pointer method is best approach because time complexity is O(n) Hash Map required addition O(n) space complexity.
I read this answer in Data structure book by Narasimha Karamanchi.
We can use Floyd cycle finding algorithm, also known as tortoise and hare algorithm. In this, two pointers are used; one (say slowPtr) is advanced by a single node, and another (say fastPtr) is advanced by two nodes. If any loop is present in the single linked list, they both will surely meet at some point.
struct Node{
int data;
struct Node *next;
}
// program to find the begin of the loop
int detectLoopandFindBegin(struct Node *head){
struct Node *slowPtr = head, *fastPtr = head;
int loopExists = 0;
// this while loop will find if there exists a loop or not.
while(slowPtr && fastPtr && fastPtr->next){
slowPtr = slowPtr->next;
fastPtr = fastPtr->next->next;
if(slowPtr == fastPtr)
loopExists = 1;
break;
}
If there exists any loop then we point one of the pointers to the head and now advance both of them by single node. The node at which they will meet will be the start node of the loop in the single linked list.
if(loopExists){
slowPtr = head;
while(slowPtr != fastPtr){
fastPtr = fastPtr->next;
slowPtr = slowPtr->next;
}
return slowPtr;
}
return NULL;
}
For the most part all the previous answers are correct but here is a simplified version of the logic with visual & code (for Python 3.7)
The logic is very simple as others explained it. I'm gonna create Tortoise/slow and Hare/fast. If we move two pointers with different speed then eventually fast will meet the slow !! you can also think of this as two runners in a tack circular field. If the fast runner keeps going in circle then it will meet/pass the slow runner.
So, we will move Tortoise/slow pointer with speed 1 for each iteration while we keep incrementing or move the Hare/fast pointer with speed of 2. Once they meet we know there is a cycle. This is also known as Floyd's cycle-finding algorithm
Here is the Python code that does this (notice has_cycle method is the main part):
#!/usr/bin/env python3
class Node:
def __init__(self, data = None):
self.data = data
self.next = None
def strnode (self):
print(self.data)
class LinkedList:
def __init__(self):
self.numnodes = 0
self.head = None
def insertLast(self, data):
newnode = Node(data)
newnode.next = None
if self.head == None:
self.head = newnode
return
lnode = self.head
while lnode.next != None :
lnode = lnode.next
lnode.next = newnode # new node is now the last node
self.numnodes += 1
def has_cycle(self):
slow, fast = self.head ,self.head
while fast != None:
if fast.next != None:
fast = fast.next.next
else:
return False
slow = slow.next
if slow == fast:
print("--slow",slow.data, "fast",fast.data)
return True
return False
linkedList = LinkedList()
linkedList.insertLast("1")
linkedList.insertLast("2")
linkedList.insertLast("3")
# Create a loop for testing
linkedList.head.next.next.next = linkedList.head;
#let's check and see !
print(linkedList.has_cycle())
Following code will find whether there is a loop in SLL and if there, will return then starting node.
int find_loop(Node *head){
Node * slow = head;
Node * fast = head;
Node * ptr1;
Node * ptr2;
int k =1, loop_found =0, i;
if(!head) return -1;
while(slow && fast && fast->next){
slow = slow->next;
/*Moving fast pointer two steps at a time */
fast = fast->next->next;
if(slow == fast){
loop_found = 1;
break;
}
}
if(loop_found){
/* We have detected a loop */
/*Let's count the number of nodes in this loop node */
ptr1 = fast;
while(ptr1 && ptr1->next != slow){
ptr1 = ptr1->next;
k++;
}
/* Now move the other pointer by K nodes */
ptr2 = head;
ptr1 = head;
for(i=0; i<k; i++){
ptr2 = ptr2->next;
}
/* Now if we move ptr1 and ptr2 with same speed they will meet at start of loop */
while(ptr1 != ptr2){
ptr1 = ptr1->next;
ptr2 = ptr2->next;
}
return ptr1->data;
}
boolean hasLoop(Node *head)
{
Node *current = head;
Node *check = null;
int firstPtr = 0;
int secondPtr = 2;
do {
if (check == current) return true;
if (firstPtr >= secondPtr){
check = current;
firstPtr = 0;
secondPtr= 2*secondPtr;
}
firstPtr ++;
} while (current = current->next());
return false;
}
Another O(n) solution.
As I viewed the selected answer, I tried a couple of examples and found that:
If (A1,B1), (A2,B2) ... (AN, BN) are the traversals of pointers A and B
where A steps 1 element and B steps 2 elements, and, Ai and Bj are the nodes traversed by A and B, and AN=BN.
Then, the node from where the loop starts is Ak, where k = floor(N/2).
ok - I ran into this in an interview yesterday - no reference material available and I came up with a very different answer (while driving home of course...) Since the linked lists are NORMALLY (not always I admit) allocated using malloc logic then we know that the granularity of the allocations is known. On most systems this is 8 bytes - this means that the bottom 3 bits are always zeros. Consider - if we place the linked list in a class to control access and use a mask of 0x0E ored into the next address then we can use the lower 3 bits to store a break crumb Thus we can write a method that will store our last breadcrumb - say 1 or 2 - and alternate them. Our method that checks for a loop can then step through each node (using our next method) and check if the next address contains the current breadcrumb - if it does we have a loop - if it does not then we would mask the lower 3 bits and add our current breadcrumb. The breadcrumb checking algorithm would have to be single threaded as you could not run two of them at once but it would allow other threads to access the list async - with the usual caveats about adding/deleting nodes.
What do you think? If others feel this is a valid solution I can write up the sample class ... Just think sometimes a fresh approach is good and am always willing to be told I have just missed the point... Thanks All Mark
Another solution
Detecting a Loop:
create a list
loop through the linkedlist and keep on adding the node to the list.
If the Node is already present in the List, we have a loop.
Removal of loop:
In the Step#2 above, while loop through the linked list we are also keep track of the previous node.
Once we detect the loop in Step#3, set previous node's next value to NULL
#code
def detect_remove_loop(head)
cur_node = head
node_list = []
while cur_node.next is not None:
prev_node = cur_node
cur_node = cur_node.next
if cur_node not in node_list:
node_list.append(cur_node)
else:
print('Loop Detected')
prev_node.next = None
return
print('No Loop detected')
Firstly, Create a Node
struct Node {
int data;
struct Node* next;
};
Initialize head pointer globally
Struct Node* head = NULL;
Insert some data in Linked List
void insert(int newdata){
Node* newNode = new Node();
newNode->data = newdata;
newNode->next = head;
head = newNode;
}
Create a function detectLoop()
void detectLoop(){
if (head == NULL || head->next == NULL){
cout<< "\nNo Lopp Found in Linked List";
}
else{
Node* slow = head;
Node* fast = head->next;
while((fast && fast->next) && fast != NULL){
if(fast == slow){
cout<<"Loop Found";
break;
}
fast = fast->next->next;
slow = slow->next;
}
if(fast->next == NULL){
cout<<"Not Found";
}
}
}
Call the function from main()
int main()
{
insert(4);
insert(3);
insert(2);
insert(1);
//Created a Loop for Testing, Comment the next line to check the unloop linkedlist
head->next->next->next->next = head->next;
detectLoop();
//If you uncomment the display function and make a loop in linked list and then run the code you will find infinite loop
//display();
}
bool FindLoop(struct node *head)
{
struct node *current1,*current2;
current1=head;
current2=head;
while(current1!=NULL && current2!= NULL && current2->next!= NULL)
{
current1=current1->next;
current2=current2->next->next;
if(current1==current2)
{
return true;
}
}
return false;
}
A quite different method:-
Reverse the linked list.
While reversing if you reach the head again then there is a loop in the list,
if you get NULL then there is no loop.
The total time complexity is O(n)
I was looking at the code below from stanford library:
void recursiveReverse(struct node** head_ref)
{
struct node* first;
struct node* rest;
/* empty list */
if (*head_ref == NULL)
return;
/* suppose first = {1, 2, 3}, rest = {2, 3} */
first = *head_ref;
rest = first->next;
/* List has only one node */
if (rest == NULL)
return;
/* put the first element on the end of the list */
recursiveReverse(&rest);
first->next->next = first;
/* tricky step -- see the diagram */
first->next = NULL;
/* fix the head pointer */
*head_ref = rest;
}
What I don't understand is in the last recursive step for e.g if list is 1-2-3-4 Now for the last recursive step first will be 1 and rest will be 2. So if you set *head_ref = rest .. that makes the head of the list 2 ?? Can someone please explain how after reversing the head of the list becomes 4 ??
Draw out a stack trace...
Intial - {1,2,3,4}
Head - 1
Rest = 2,3,4
Recurse(2,3,4)
Head = 2
Rest = 3,4
Recurse(3,4)
Head = 3
Rest = 4
Recurse (4)
Head = 4
Rest = null //Base Case Reached!! Unwind.
So now we pick up
Recurse(3,4)
Head = 3
Rest = 4
// Return picks up here
first->next->next = first;
so list is:
3,4,3
// set head to null,
null ,4,3,
//Off with his head!
4,3
Return
Now we're here
Recurse(2,3,4)
Head = 2
Rest = 3,4
Previous return leaves state as:
Head = 2 //But Head -> next is still 3! -- We haven't changed that yet..
Rest = 4,3
Head->next is 3,
Head->next->next = 2 makes the list (actually a tree now)
4->3->2
^
|
2
And chop off the head leaving
4->3->2
and return.
Similarly, do the last step which will leave
4->3->2->1
^
|
1
and chop off the head, which removes the one.
Consider the list:
1 -> 2 -> 3 -> 4 -> NULL
^ ^
| |
first rest
Where first points to the first node and rest points to the node next to first.
Since the list is not empty and list does not contain one node we make recursive call to reverse to reverse the list pointed to by rest. This is how the list looks after reversing the rest of the list:
1 -> 2 <- 3 <- 4
^ | ^
| NULL |
first rest
As seen rest now points to the reversed list which has 4 at the beginning and 2 at the end of list. The next pointer of node 2 is NULL.
Now we need to append the first node to the end of the reversed-rest list. To append anything to the end of the list we need to have access to the last node of the list. In this case we need to have access to the last node of the reversed-rest list. Look at the diagram, first -> next points to the last node reversed-rest list. Therefore first -> next -> next will be next pointer of the last node of the reversed-rest list. Now we need to make it point to first so we do:
first -> next -> next = first;
After this step the list looks like:
1 <- 2 <- 3 <- 4
^ -> ^
| |
first rest
Now the next field of the last node of the list must be NULL. But it is not the case now. The next field of the last node ( node 1) is pointing to the node before it ( node 2). To fix this we do:
first -> next = NULL;
After this the list looks like:
NULL <- 1 <- 2 <- 3 <- 4
^ ^
| |
first rest
As seen the list is now correctly reversed with rest pointing to the head of the reversed list.
We need to return the new head pointer so the that changes are reflected in the calling function. But this is a void function and head is passed as double pointer so changing the value of *head will make the calling function see the changed head:
*head = rest;
The rest isn’t 2, it’s 2 -> 3 -> 4, which gets reversed recursively. After that we set *head_ref to rest, which is now (recursively reversed!) 4 -> 3 -> 2.
The important point here is that although both first and rest have the same type, i.e. node*, they are conceptually fundamentally different: first points to one single element, while rest points to a linked list of elements. This linked list is reversed recursively before it gets assigned to *head_ref.
I recently wrote a recursive method for reversing a linked list in ruby. Here it is:
def reverse!( node_1 = #head, node_2 = #head.link )
unless node_2.link
node_2.link = node_1
#head = node_2
return node_1
else
return_node = reverse!(node_1.link, node_2.link)
return_node.link = node_1
node_1.link = nil
return node_1
end
return self
end