Given root of binary search tree and K as input, find K-th smallest element in BST.
For example, in the following BST, if k = 3, then output should be 10, and if k = 5, then output should be 14.
Method 1: Using Inorder Traversal.
Inorder traversal of BST retrieves elements of tree in the sorted order. The inorder traversal uses stack to store to be explored nodes of tree (threaded tree avoids stack and recursion for traversal, see this post). The idea is to keep track of popped elements which participate in the order statics. Hypothetical algorithm is provided below,
Time complexity: O(n) where n is total nodes in tree..
/* initialization */ pCrawl = root set initial stack element as NULL (sentinal) /* traverse upto left extreme */ while(pCrawl is valid ) stack.push(pCrawl) pCrawl = pCrawl.left /* process other nodes */ while( pCrawl = stack.pop() is valid ) stop if sufficient number of elements are popped. if( pCrawl.right is valid ) pCrawl = pCrawl.right while( pCrawl is valid ) stack.push(pCrawl) pCrawl = pCrawl.left
Method 2: Augmented Tree Data Structure.
The idea is to maintain rank of each node. We can keep track of elements in a subtree of any node while building the tree. Since we need K-th smallest element, we can maintain number of elements of left subtree in every node.
Assume that the root is having N nodes in its left subtree. If K = N + 1, root is K-th node. If K < N, we will continue our search (recursion) for the Kth smallest element in the left subtree of root. If K > N + 1, we continue our search in the right subtree for the (K – N – 1)-th smallest element. Note that we need the count of elements in left subtree only.
Time complexity: O(h) where h is height of tree.
start: if K = root.leftElement + 1 root node is the K th node. goto stop else if K > root.leftElements K = K - (root.leftElements + 1) root = root.right goto start else root = root.left goto srart stop: