## 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:

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

Appraoch:

## Lowest Common Ancestor in a Binary Search Tree.

Objective: Find the Lowest Common Ancestor of two given nodes in a Binary Search 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 Tree ( Not Binary Search Tree).

Example:

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

Appraoch:

## 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.

## Convert a Sorted Doubly Linked List to Balanced BST.

Objective: Given a sorted doubly linked list, convert it into Balanced binary search tree

Example:

Approach:

## Find the number of occurrences of a number in a given sorted array.

Objective: Given a sorted(ascending order) arrays of integers, find out the number of occurences of a number in that array

Input: A sorted array arrA[] and a number x.

Output: number of occurrences of ‘x’ in array arrA[].

Examples :

```Array - {1,2,2,2,2,2,2,2,3,4,5,5,6}
x = 2
Output : No of Occurrences of number 2 is 7
```

Approach:

## 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

Input: A binary search tree, a node x

Output: Inorder successor of node x.

Example:

## 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

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

Output: The inorder successor of node x.

Example:

## Inorder Successor in Binary Tree

Algorithms – Inorder Successor in Binary Tree

Objective: Given a Binary tree (Not 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

Input: A binary tree, a node x

Output: Inorder successor of node x.

Example:

Approach:

NOTE: since it’s not a binary search tree, we cannot use binary search technique to reach to the node. we need to travel all the nodes in order to find the node for which we need to find the inorder successor.

## 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:

## Level Order Traversal, Print each level in separate line.

Objective: Given a Binary tree , Print each level of a tree in separate line.

NOTE : This problem is very similar ” Create Linked Lists of all the nodes at each depth “

Input: A binary tree

Output: Each level of binary tree, in one line

Example:

## Rearrange Positive and Negative Numbers of Array On Each Side in O(nlogn)

Objective: Rearrange Positive and Negative Numbers of an Array so that one side you have positive numbers and other side with negative Integers without changing their respective order.

Example

``` : Input :  1 -2 3 -4 5 -6 7 -8 9 -10

ReArranged Output :  -2 -4 -6 -8 -10 1 3 5 7 9
```

Input: An array with positive and negative numbers

Output: Modified array with positive numbers and negative numbers are on each side of the array.

Approach:

Method 1. One naive approach is to have another array of same size and navigate the input array and one scan place all the negative numbers to the new array and in second scan place all the positive numbers to new array. Here the Space complexity will be O(n). We have a better solution which can solve this in O(1) space.

Method 2: Divide and Conquer

## Find The Longest Sequence Of Prefix Shared By All The Words In A String

Objective: Write an algorithm to find The Longest Sequence Of Prefix Shared By All The Words In A String. This interesting problem was asked in the Google interview for software engineer. This problem is bit tricky, it looks difficult but actually it is simple problem.

Input: A String

Output: The longest sequence of prefix common in all the words in a string

Example:

“Bedroom BedClock BedTable BedWall” => “Bed”

Approach: