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

Here's just an outline of the idea:

In a BST, the left subtree of node T contains only elements smaller than the value stored in T. If k is smaller than the number of elements in the left subtree, the kth smallest element must belong to the left subtree. Otherwise, if k is larger, then the kth smallest element is in the right subtree.

We can augment the BST to have each node in it store the number of elements in its left subtree (assume that the left subtree of a given node includes that node). With this piece of information, it is simple to traverse the tree by repeatedly asking for the number of elements in the left subtree, to decide whether to do recurse into the left or right subtree.

Now, suppose we are at node T:

1. If k == num_elements(left subtree of T), then the answer we're looking for is the value in node T.
2. If 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. If 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.

Complexity analysis:

This takes O(depth of node) time, which is O(log n) in the worst case on a balanced BST, or O(log n) on average for a random BST.

A BST requires O(n) storage, and it takes another O(n) to store the information about the number of elements. All BST operations take O(depth of node) time, and it takes O(depth of node) extra time to maintain the "number of elements" information for insertion, deletion or rotation of nodes. Therefore, storing information about the number of elements in the left subtree keeps the space and time complexity of a BST.

You can use iterative inorder traversal: http://en.wikipedia.org/wiki/Tree_traversal#Iterative_Traversal with a simple check for kth element after poping a node out of the stack.

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.

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.

While this is definitely not the optimal solution to the problem, it is another potential solution which I thought some people might find interesting:

/**
 * Treat the bst as a sorted list in descending order and find the element 
 * in position k.
 *
 * Time complexity BigO ( n^2 )
 *
 * 2n + sum( 1 * n/2 + 2 * n/4 + ... ( 2^n-1) * n/n ) = 
 * 2n + sigma a=1 to n ( (2^(a-1)) * n / 2^a ) = 2n + n(n-1)/4
 *
 * @param t The root of the binary search tree.
 * @param k The position of the element to find.
 * @return The value of the element at position k.
 */
public static int kElement2( Node t, int k ) {
    int treeSize = sizeOfTree( t );

    return kElement2( t, k, treeSize, 0 ).intValue();
}

/**
 * Find the value at position k in the bst by doing an in-order traversal 
 * of the tree and mapping the ascending order index to the descending order 
 * index.
 *
 *
 * @param t Root of the bst to search in.
 * @param k Index of the element being searched for.
 * @param treeSize Size of the entire bst.
 * @param count The number of node already visited.
 * @return Either the value of the kth node, or Double.POSITIVE_INFINITY if 
 *         not found in this sub-tree.
 */
private static Double kElement2( Node t, int k, int treeSize, int count ) {
    // Double.POSITIVE_INFINITY is a marker value indicating that the kth 
    // element wasn't found in this sub-tree.
    if ( t == null )
        return Double.POSITIVE_INFINITY;

    Double kea = kElement2( t.getLeftSon(), k, treeSize, count );

    if ( kea != Double.POSITIVE_INFINITY )
        return kea;

    // The index of the current node.
    count += 1 + sizeOfTree( t.getLeftSon() );

    // Given any index from the ascending in order traversal of the bst, 
    // treeSize + 1 - index gives the
    // corresponding index in the descending order list.
    if ( ( treeSize + 1 - count ) == k )
        return (double)t.getNumber();

    return kElement2( t.getRightSon(), k, treeSize, count );
}

Well we can simply use the in order traversal and push the visited element onto a stack. pop k number of times, to get the answer.

we can also stop after k elements

Python Solution Time Complexity : O(n) Space Complexity : O(1)

Idea is to use Morris Inorder Traversal

class Solution(object):
def inorderTraversal(self, current , k ):
    while(current is not None):    #This Means we have reached Right Most Node i.e end of LDR traversal

        if(current.left is not None):  #If Left Exists traverse Left First
            pre = current.left   #Goal is to find the node which will be just before the current node i.e predecessor of current node, let's say current is D in LDR goal is to find L here
            while(pre.right is not None and pre.right != current ): #Find predecesor here
                pre = pre.right
            if(pre.right is None):  #In this case predecessor is found , now link this predecessor to current so that there is a path and current is not lost
                pre.right = current
                current = current.left
            else:                   #This means we have traverse all nodes left to current so in LDR traversal of L is done
                k -= 1
                if(k == 0):
                    return current.val
                pre.right = None       #Remove the link tree restored to original here 
                current = current.right
        else:               #In LDR  LD traversal is done move to R 
            k -= 1
            if(k == 0):
                return current.val
            current = current.right

    return 0

def kthSmallest(self, root, k):
    return self.inorderTraversal( root , k  )

public int kthSmallest(TreeNode root, int k) {
     
    LinkedList<TreeNode> stack = new LinkedList<TreeNode>();

    while (true) {
      while (root != null) {
        stack.push(root);
        root = root.left;
      }
      root = stack.pop();
      k = k - 1;
      if (k == 0) return root.val;
      root = root.right;
    }

}     

Here are the steps:

1.Add a field to each node indicating the size of the tree it roots. This supports operation in O(logN) average time.

2.To save space, one field indicating the size of a node it roots is enough. We don't need to save both the left subtree and right subtree size.

3.Do an inorder traversal until LeftTree == K, LeftTree = Size(T->Left) + 1.

4.Here is the sample code:

int Size(SearchTree T)
{
    if(T == NULL) return 0;
    return T->Size;
}
Position KthSmallest(SearchTree T, int K)
{
    if(T == NULL) return NULL;

    int LeftTree;
    LeftTree = Size(T->Left) + 1;

    if(LeftTree == K) return T;

    if(LeftTree > K){ 
        T = KthSmallest(T->Left, K); 
    }else if(LeftTree < K){ 
        T = KthSmallest(T->Right, K - LeftTree);
    }   

    return T;
}

5.Similarly, we can also get the KthLargest function.

 public int printInorder(Node node, int k) 
    { 
        if (node == null || k <= 0) //Stop traversing once you found the k-th smallest element
            return k; 

        /* first recur on left child */
        k = printInorder(node.left, k); 

        k--;
        if(k == 0) {  
            System.out.print(node.key);
        }

        /* now recur on right child */
        return printInorder(node.right, k);
    } 

This java recursive algorithm, stops the recursion once the k-th smallest element is found.

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.