Median of a BST in O(1) space
You are given a binary search tree of integers with N nodes. Your task is to find the median of the given BST.
The median of a BST is the middle element when all the data nodes are written in ascending order. In other words-
• If ‘N’ is even, then the median is the integral average of N/2-th and (N/2+1)-th nodes when nodes are written in ascending order. • If ‘N’ is odd, then the median is the (N+1)/2-th node when nodes are written in ascending order.
A binary search tree (BST) is a binary tree data structure which has the following properties.
• The left subtree of a node contains only nodes with data less than the node’s data. • The right subtree of a node contains only nodes with data greater than the node’s data. • Both the left and right subtrees must also be binary search trees.
Try to do this problem in O(1) space and linear time.
The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then the test cases follow. The first line of input contains the elements of the tree in the level order form separated by a single space. If any node does not have a left or right child, take -1 in its place. Refer to the example below. Example: Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place. For example, the input for the tree depicted in the below image would be :
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1 Explanation : Level 1 : The root node of the tree is 1 Level 2 : Left child of 1 = 2 Right child of 1 = 3 Level 3 : Left child of 2 = 4 Right child of 2 = null (-1) Left child of 3 = 5 Right child of 3 = 6 Level 4 : Left child of 4 = null (-1) Right child of 4 = 7 Left child of 5 = null (-1) Right child of 5 = null (-1) Left child of 6 = null (-1) Right child of 6 = null (-1) Level 5 : Left child of 7 = null (-1) Right child of 7 = null (-1) The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on. The input ends when all nodes at the last level are null (-1).
The above format was just to provide clarity on how the input is formed for a given tree. The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as: 1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
Output Format :
For each test case, print a single line containing the median of the BST. The output for each test case will be printed in a new line.
You do not need to print anything; it has already been taken care of. Just implement the given function.
1 <= T <= 100 1 <= N <= 5000 0 <= Data <= 10^6 and Data!=-1 Data will be distinct. Time Limit: 1sec
The key observation here is that the inorder traversal of the BST gives all its elements in ascending order. We will first find the size of the BST using one inorder traversal. Using another inorder traversal we can return the (N + 1) / 2-th node or integer average of N / 2-th and N / 2 + 1-th nodes depending on whether ‘N’ is odd or even.
The algorithm for the Inorder Traversal will be-
- Initialize the current Node ‘CURRNODE’ as the root of the BST.
- While ‘CURRNODE’ is not equal to NULL check if ‘CURRNODE’ has a left child.
- If ‘CURRNODE’ does not have a left child, then it is the next node in the inorder traversal of the BST. So, update ‘CURRNODE’ and point it to the node on the right of ‘CURRNODE’.
- Else we do the following-
- We need to find the inorder predecessor of the ‘CURRNODE’. So we find the rightmost node in ‘CURRNODE’ left subtree or node whose right child is equal to ‘CURRNODE’ if there is no left subtree. We can use another pointer ‘PREV’ for this.
- If we found the node whose right child is equal to ‘CURRNODE’.
- Then ‘CURRNODE’ is the next node in the inorder traversal of BST. Next, we go to the right child, that is we do CURRNODE = CURRNODE -> right.
- Else, we do the following-
- We make ‘CURRNODE’ as the right child of its inorder predecessor. So, we make ‘CURRNODE’ as the right child of the rightmost node we found.
- We go to the left child i.e CURRNODE = CURRNODE -> left.