CODEWITHSUNDEEP
c#data-structuresbinary-treebinary-search-tree
Mostafa Ghorbani
Mostafa Ghorbani

Reputation: 445

binary search tree (BST) find method in C#

how can I Write a code with these requirements? this is the question: check if there is a node with key K in the tree T, and if so, return a reference to this node. If the tree is empty, report that the node was not found and stop. Otherwise, compare K with the key value of the root node X.

This is what i found:

public GenBT<T> Find(GenBT<T> k, T inf){ 

if (k == null) return null; 

else 

switch (inf.CompareTo(k.inf)){ 

case 1: return Find(k.RT, inf); 

case -1: return Find(k.LT, inf); 

case 0: return k; 

default: return null;} 

};

but I also found that if I want to search or find in BST I have to use a code like this :

struct node* search(int data){
   struct node *current = root;
   printf("Visiting elements: ");
    
   while(current->data != data){
    
      if(current != NULL) {
         printf("%d ",current->data);
            
         //go to left tree
         if(current->data > data){
            current = current->leftChild;
         }  //else go to right tree
         else {                
            current = current->rightChild;
         }
            
         //not found
         if(current == NULL){
            return NULL;
         }
      }         
   }
   
   return current;
}

These two look very different and I don't know which way is right and in general what is the right way of solving this question

Upvotes: 0

Views: 1239

Answers (1)

JonasH
JonasH

Reputation: 36371

To transform the recursive solution to an iterative one we need to insert a loop. In the recursive case we change the node parameter for each recursion. To do the same in the iterative case we simply create a variable that is updated instead of doing a recursion.

Changed naming to make a clearer and compilable example. Note also that CompareTo can return any number, not just 0, 1, -1. So a switch is insufficient:

    public class Node<T>
    {
        public Node<T> Left { get; }
        public Node<T> Right { get; }
        public T Value { get; }

        public Node(Node<T> left, Node<T> right, T value)
            => (Left, Right, Value) = (left, right, value);
    }

    public static Node<T> Find<T>(Node<T> root, T target) where T : IComparable<T>
    {
        var current = root;
        while (current != null)
        {
            var comparison = target.CompareTo(current.Value);
            if (comparison > 0) 
                current = current.Right;
            else if (comparison < 0)
                current = current.Left;
            else
                return current;

        }
        return null;
    }

See also how to transform a recursive method to an iterative one in a more generic way. Note that the example uses a stack, In your case a stack is not needed, since you only process one branch of the tree.

Upvotes: 2

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