Data structures

Question : Write a function to print the lowest common ancestor of two nodes in a binary search tree.  Lowest common ancestor of two nodes x and y in a binary tree is the lowest node that has both x and y as descendants. Here lowest common ancestor of 1 and 7 is 3. LCA of 13 and 7 is root - 8. And LCA of 6 and 7 is 6 itself. The program to find gets complicated for an ordinary binary tree. But for a binary search tree, it is quite simple. As we see from the diagram above, the paths to 1 and 4 are common till the node 3. And at 3 they branch in different directions. So 3 is our LCA. That is lowest common ancestor is the node where the paths to x and y from root deviate. As long as they branch in same direction, we continue to traverse. When they branch in different directions, that is the lowest common ancestor. So let us write a simple algorithm, set temp=root if temp->val >x and temp->val>y temp = temp->left else if temp->val<x and ...

Question : Write a function for in-order traversal of nodes in the range x to y from a binary search tree. This is quite a simple function. As a first solution we can just traverse our binary search tree in inorder and display only the nodes which are in the range x to y. But if the current node has a value less than x, do we have to traverse its left subtree? No. Because all the nodes in left subtree will be smaller than x. Similarly if the current node has a key value more than y, we need not visit its right subtree. Now we are ready to write our algorithm.     if nd is NOT NULL  if nd->val >=x then visit all the nodes of left subtree of nd recursively display nd->val if nd->val <y then visit all the nodes of right subtree of nd recursively  That's all. We have our function ready. void in_order_middle (NODEPTR nd, int x, int y) { if (nd) { if (nd -> val >= x) in_order_middle(nd...

How do you merge two binary search trees? I googled about the solutions. Most solutions told me to convert both trees into linked lists. Merge the lists. Then create a tree from the elements of the list. But why lists? Why can't we store the elements in an array? Because if the data of the tree is larger - not just integer keys, array manipulation becomes difficult. But again, we need not convert both the trees into lists. We can convert one tree into list - a doubly linked list. Then insert the elements of this list into the other tree. I tried this approach. To convert a tree into a sorted doubly linked list Create a doubly linked list. Let the prev and next links of nodes in this list be called left and right respectively. This way we can directly use the binary tree nodes in the list. Use a static variable previousnode  call the function recursively for left child of current node. link current node to the previousnode set next pointer of previousnode to curre...

I came across this question in the net. Given a binary search tree, convert this into a sorted doubly linked list. Do not create any nodes, but just modify the links. Just like a DLL, binary tree node also has two links. But they are named left and right. What if we assume left is previous link and right is next link? OK, we don't even have to modify the structure of the nodes. Next important part is "sorted". We need a sorted DLL. Don't you get the idea now. Which traversal visits the nodes in ascending order or gives sorted output? In order traversal! That's right. So we need to visit the nodes in in-order traversal, store previous node in a static variable, and link this previous node to the next node visited using previous (left) and next(right) links. Now how do we determine the head of DLL? It should be the tree-minimum. So we can find tree minimum and set this as head. Or we can use in order traversal and set the node which has no previous node visited as h...

This is one of the interview questions. Given a binary tree, find out whether it is a binary search tree. A binary search tree is an ordered binary tree which satisfies  binary search tree property, which states that the key in each node must be greater than or equal to any key stored in the left sub-tree, and less than or equal to any key stored in the right sub-tree. A binary Search Tree   This diagram shows a BST, because each node has key value larger than left subtree and smaller than right subtree. Binary Tree but not a Binary search tree This is not a binary search tree because key value of left child of root is larger than 2. And key value of node 7 is has a right child whose value is lesser than 7. How to write a function If the in order traversal output does not give key values in ascending order, then the tree is not a binary search tree. But inorder is a recursive function. So we can not directly compare values. We need to use a global variable to store the valu...

In an earlier post , we have seen how to add a new node to a binary tree. Binary Tree Terminology: Tree is a non-linear data structure where every node has multiple branches A binary tree is a tree where each node has maximum 2 branches Each node branching out is called child node  The starting node of the tree is called root The two children of binary tree are left child and right child. The child node along with its branches and sub-branches are called sub-tree.  A node has two sub-trees - left subtree and right subtree A node which has no child nodes and hence no subtrees is called a leaf node Image from : http://msoe.us/taylor Normally when we talk about binary tree, we refer to binary search tree which is ordered tree. In a binary search tree, every node to the right of a given node will have larger value than the given node. And every node to the left of given node will have value smaller than given node. That is to say the left subtree of an...