Find kth smallest element in a binary search tree in Optimum way

Question

Hi,

I need to find the kth smallest element in the binary search tree without using any static/global variable. How to achieve it efficiently? The solution that I have in my mind is doing the operation in O(n), the worst case since I am planning to do an inorder traversal of the entire tree. But deep down I feel that I am not using the BST property here. Is my assumptive solution correct or is there a better one available ?

Answer 1

For a binary search tree, an inorder traversal will return elements ... in order.

Just do an inorder traversal and stop after traversing k elements.

O(1) for constant values of k.

Answer 2

Here's just an outline of the idea:

Let each node in the BST have a field that returns the number of elements in its left and right subtree. Let the left subtree of node T contain only elements smaller than T and the right subtree only elements larger than or equal to T.

Now, suppose we are at node T:

1. k == num_elements(left subtree of T), then the answer we're looking for is the value in node T
2. k > num_elements(left subtree of T) then obviously we can ignore the left subtree, because those elements will also be smaller than the kth smallest. So, we reduce the problem to finding the k - num_elements(left subtree of T) smallest element of the right subtree.
3. k < num_elements(left subtree of T), then the kth smallest is somewhere in the left subtree, so we reduce the problem to finding the kth smallest element in the left subtree.

This is O(log N) on average (assuming a balanced tree).

Answer 3

Given just a plain binary search tree, about all you can do is start from the smallest, and traverse upward to find the right node.

If you're going to do this very often, you can add an attribute to each node signifying how many nodes are in its left sub-tree. Using that, you can descend the tree directly to the correct node.

Answer 4

If you do a search on "Order Statistics" you will find what you need.

Answer 5

For not balanced searching tree, it takes O(n).

For balanced searching tree, it takes O(k + log n) in the worst case but just O(k) in Amortized sense.

Having and managing the extra integer for every node: the size of the sub-tree gives O(log n) time complexity. Such balanced searching tree is usually called RankTree.

In general, there are solutions (based not on tree).

Regards.

Answer 6

This is what I though and it works. It will run in o(log n )

public static int FindkThSmallestElemet(Node root, int k)
    {
        int count = 0;
        Node current = root;

        while (current != null)
        {
            count++;
            current = current.left;
        }
        current = root;

        while (current != null)
        {
            if (count == k)
                return current.data;
            else
            {
                current = current.left;
                count--;
            }
        }

        return -1;


    } // end of function FindkThSmallestElemet

Answer 7

public int ReturnKthSmallestElement1(int k)
    {
        Node node = Root;

        int count = k;

        int sizeOfLeftSubtree = 0;

        while(node != null)
        {

            sizeOfLeftSubtree = node.SizeOfLeftSubtree();

            if (sizeOfLeftSubtree + 1 == count)
                return node.Value;
            else if (sizeOfLeftSubtree < count)
            {
                node = node.Right;
                count -= sizeOfLeftSubtree+1;
            }
            else
            {
                node = node.Left;
            }
        }

        return -1;
    }

this is my implementation in C# based on the algorithm above just thought I'd post it so people can understand better it works for me

thank you IVlad

Answer 8

Given a Binary Search Tree, write a program to print the kth smallest element without using any static/global variable. You can't pass the value k to any function also.

Answer 9

This works well: status : is the array which holds whether element is found. k : is kth element to be found. count : keeps track of number of nodes traversed during the tree traversal.

int kth(struct tree* node, int* status, int k, int count)
{
    if (!node) return count;
    count = kth(node->lft, status, k, count);  
    if( status[1] ) return status[0];
    if (count == k) { 
        status[0] = node->val;
        status[1] = 1;
        return status[0];
    }
    count = kth(node->rgt, status, k, count+1);
    if( status[1] ) return status[0];
    return count;
}

Answer 10

Well here is my 2 cents...

int numBSTnodes(const Node* pNode){
     if(pNode == NULL) return 0;

     return (numBSTnodes(pNode->left)+numBSTnodes(pNode->right)+1);
}


//This function will find Kth smallest element
Node* findKthSmallestBSTelement(Node* root, int k){
     Node* pTrav = root;
     while(k > 0){
         int numNodes = numBSTnodes(pTrav->left);
         if(numNodes >= k){
              pTrav = pTrav->left;
         }
         else{
              //subtract left tree nodes and root count from 'k'
              k -= (numBSTnodes(pTrav->left) + 1);

              if(k == 0) return pTrav;

              pTrav = pTrav->right;
        }

        return NULL;
 }