Data structures

Question : Write a function to find inorder successor of a given node in a binary  search tree. Our inclination would be to write tree minimum of right subtree. Because we already use this mechanism while deleting a node with both subtrees. But what we fail to notice is that tree minimum of right subtree would be present if there IS a right subtree. What if the node has no right child? What if it is a leaf node? Which nodes are successors of these nodes? And how do we find them? The inorder successor of 3 can be found out as tree minimum of right subtree, which is 4. But what about successors of 4? Or 7? Or successor 13? One method would be to store parent link in each node. And travel up finding parent nodes until the parent node has a value greater than given node. But that needs modification of the structure. Another method is to start from root node and search for a given value and store node with larger value as successor and terminate when we come across a node w...

Question : Write a function for BFS traversal of a binary tree. Bread first traversal (also called level order traversal) is a traversal method where you visit the siblings of a node before you visit its descendants. The other method of traversal is called depth first traversal(DFS) where you visit the descendants of a node before you visit its siblings. In-order, pre-order and post-order are all DFS traversal methods. In BFS, first you visit all the nodes at level 0, then you visit all the nodes at level 1, then you visit nodes at level 2 etc. For the diagram shown above, the BFS output should be 8 3 10 1 6 14  4 7 13  ( there is no newline between levels) To write a code for BFS, we need to take the help of another data structure - a queue. To start with we insert this root to the queue. Then as long as queue is not empty, we remove a node from the queue, visit this node (display its value) and enqueue its both child nodes.  Here is C function f...

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...

Question : Write a function to mirror a binary tree. The mirror tree will have a left subtree and right subtree swapped. But this has to be done for each and every node. So to convert a tree to its mirror, we can write an algorithm like this               1) mirror left subtree recursively               2) mirror right subtree recursively               3) swap left and right children of the current node. Here is the function for it. void mirror_tree (NODEPTR root) { if (root != NULL ) { NODEPTR temp = root -> left; root -> left = root -> right; root -> right = temp; mirror_tree(root -> left); mirror_tree(root -> right); } }

A binary tree is a non-linear data structure where every node has maximum two branches - left subtree and right subtree. An empty node is also a binary tree. In earlier posts we have seen how to insert a node to binary tree , how to traverse a tree and how to delete nodes from tree. The functions look almost similar in C++. Except that all these functions and root pointer are all parts of the class binary tree. That is a good thing and a bad thing. If root is a data member (so private), how do we use root as a parameter in all the recursive functions - insert, delete, inorder etc? Easy answer would be don't use recursion. Anyway no one like them. So we can write insert and delete functions without using recursion and even search function. But traversal functions have to be recursive. Ok, let us write a function to return root of the tree and use it as first parameter in all traversal functions. Here is the complete class. You can download this code along with t...

Question : Write a function to test if two binary trees are equal. The trees are equal only if the values are equal and also the structures are equal. Equal trees       89                                89   12      100                    12       100 2                                 2 Not equal trees         89                          ...

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...