Construct Binary Search Tree from a given Preorder Traversal using Recursion

Objective: – Given a preorder traversal, construct BST from that. Input: Preorder traversal Similar Problem : This problem is similar to the – Construct Binary Search Tree from a given Preorder Traversal Using Stack (Without Recursion) Approach: Solution to the problem is similar to isBST Max-Min Solution. “Your root value can have any value between … Read more

Construct a Binary Tree from Given Inorder and Depth-First-Search.

Objective: Given a inorder and pre­order tra­ver­sal, con­struct a binary tree from that.

Input: Inorder traversal and Depth-First-Search.

Approach:

int[] inOrder = { 8, 4, 2, 5, 1, 6, 3, 7, 9 };

int[] DFS = { 1, 2, 4, 8, 5, 3, 6, 7, 9 };

Read more

Inorder Predecessor and Successor in Binary Search Tree

Objective: Given a Binary Search Tree, Find predecessor and Successor of a given node.

What is Predecessor and Successor :

When you do the inorder traversal of a binary tree, the neighbors of given node are called Predecessor(the node lies behind of given node) and Successor (the node lies ahead of given node).

Example:

Inorder Predecessor and Successor in Binary Search Tree

Approach:

Read more

Rearrange Positive and Negative Elements at Alternate Positions in an Array In O(1) Extra Space

Objective: Given an array arrA[] which has negative and positive elements, rearrange the array in such a manner that positive and negative elements occupy the alternate positions and if there are extra positive or negative elements are left then append it to the end.

Examples:

int[] arrA = { 1, 2, -3, -4, -5, 6, -7, -8, 9, 10, -11, -12, -13, 14 };
Output: -13 9 -3 10 -5 6 -7 2 -12 1 -11 14 -4 -8

Naive Solution: Take the extra space of O(n), do the linear scan twice and fill the positive and negative elements at alternate positions.

But the problem becomes interesting when you have asked not to use any extra space, ask to solve in O(1).

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

Use Quick sort technique.

  • Take the pivot element as 0 and do the first round of Quick Sort.
  • After above step you will have all the negative elements on left and all the positive elements on the right.

Read more

Find All Elements in an Array which appears more than N/K times, N is Array Size and k is a Number.

Objective: Given an array of size of N and number k. Find all elements in an Array which appears more than N/K times.

Input: Array [] and number k.

Example:

int[] arrA = { 2, 2, 4, 4, 3, 5, 3, 4, 4, 6, 4, 3, 3, 8 };

K = 4

N/k = 14/4 = 3

Output will be [3,4] they appear 5, 4 times respectively.

Approach:

Read more

Print All The Nodes Which are X distance from the Given Node

Objective: Given Binary Tree, Print All The Nodes Which are X distance from the Given Node.

Example :

Approach:

Quite Tricky solution, i will explain using the example given in the picture.

Read more

Construct a binary tree from given Inorder and Postorder Traversal

Objective: Given a inorder and postorder traversal, write an algorithm to construct a binary tree from that. This problem was asked in the Microsoft coding competition.

Input: Inorder and postorder traversals

Similar Problems: Make a Binary Tree from Given Inorder and Preorder Traveral.

Appraoch:

int[] inOrder = { 4, 2, 5, 1, 6, 3, 7 };

int[] postOrder = { 4, 5, 2, 6, 7, 3, 1 };.

Read more

Lowest Common Ancestor in a Binary Tree (Not Binary Search Tree).

Objective: Find the Lowest Common Ancestor of two given nodes in a Binary Tree

What is Lowest Common Ancestor

In a given binary tree, The lowest common ancestor of two nodes n1 and n2 will be a node X such that node X will be the lowest node who has n1 and n2 as its descendants.

Similar Problem: Lowest Common Ancestor in a Binary Search Tree.

Example:

Lowest Common Ancestor Example

Input: A binary Tree and two nodes n1 and n2.

Appraoch:

Read more

Make a Binary Tree from Given Inorder and Preorder Traveral.

Objective: Given a inorder and preorder traversal, construct a binary tree from that.

Input: Inorder and preorder traversals

Similar Problem: Construct a binary tree from given Inorder and Postorder Traversal

Approach:

int [] inOrder = {2,5,6,10,12,14,15};

int [] preOrder = {10,5,2,6,14,12,15};

  • First element in preorder[] will be the root of the tree, here its 10.
  • Now the search element 10 in inorder[], say you find it at position i, once you find it, make note of elements which are left to i (this will construct the leftsubtree) and elements which are right to i ( this will construct the rightSubtree).
  • See this step above and recursively construct left subtree and link it root.left and recursively construct right subtree and link it root.right.

Read more

Inorder Successor in Binary Search Tree without Using Parent link

Objective: Given a Binary Search tree, find the inorder successor of a node.

What is Inorder Successor: Inorder successor of a node is the next node in the inorder traversal of the tree. For the last node in a tree, inorder successor will be NULL

Similar Problems :
Inorder Successor in Binary Search Tree with parent link
Inorder Successor in Binary Tree

Input: A binary search tree, a node x

Output: Inorder successor of node x.

Read more

Inorder Successor in Binary Search Tree Using Parent link

Algorithms – Inorder Successor in Binary Search Tree Using Parent link

Objective: Given a Binary Search tree in which every node has a link to its parent, find the inorder successor of a node.

What is Inorder Successor: Inorder successor of a node is the next node in the inorder traversal of the tree. For the last node in a tree, inorder successor will be NULL

Similar Problems :
Inorder Successor in Binary Search Tree without parent link
Inorder Successor in Binary Tree

Input: A binary search tree with nodes liked to its parents, a node x

Output: The inorder successor of node x.

Read more

In a Binary Tree, Create Linked Lists of all the nodes at each depth.

Objective: Given a Binary tree create Linked Lists of all the nodes at each depth , say if the tree has height k then create k linked lists.

NOTE : This problem is very similar “Print binary tree, each level in one line

Input: A binary tree

Output: K linked lists if the height of tree is k. Each linked list will have all the nodes of each level.

Example:

Read more