Pair sum in a BST

Easy
0/40
Average time to solve is 15m
16 upvotes
Asked in companies
MicrosoftAmazonSamsung

Problem statement

You are given a binary search tree and an integer ‘S’. Your task is to find all the pairs of nodes in the BST which sum to the value ‘S’. If no such pair exists, then print -1 - 1.

Note:

You can use extra space of the order of not more than O(log n).

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.

Note:

1. All the elements of the Binary Search Tree are unique.
2. You can’t use the same node value/element of BST twice.
Detailed explanation ( Input/output format, Notes, Images )

Input Format:

The first line of input contains an integer ‘T’, which denotes the number of test cases. Then each test case  follows. 

The first line of every test case contains elements of the Binary Search Tree 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.

The second line of every test case contains a single integer ‘K’ which denotes the target value.

For example:

The input for the tree depicted in the below image would be :


20 10 35 5 15 30 42 -1 13 -1 -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).

Note:

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:

Output contains each pair of integers (pair elements separated by space). The order of different pairs, to be printed, should be in sorted order.

Note:

You don’t need to print anything, it has already been taken care of. Just implement the given function.

Constraints:

1 <= T <= 5    
1 <= N <= 10 ^ 3
0 <= DATA <= 10 ^ 9
1 <= S <= 10 ^ 9 

Where ‘DATA’ denotes the value of each node in the given tree.


Time limit: 1 sec
Sample Input 1:
2 
8 5 10 2 6 -1 -1 -1 -1 -1 7 -1 -1
13
8 5 10 2 6 -1 -1 -1 -1 -1 7 -1 -1
19
Sample Output 1:
5 8 6 7
-1 -1
Explanation for sample input 1:
For the first test case, the nodes with values 8, 5 and 6, 7 give a sum equal to the given target 13. Therefore, the output will be 5, 8 and 6, 7 in separate lines.

For the second test case, there are no two elements in the given BST such that their sum is equal to the given target ‘K’ = 19. Therefore output will be nothing.

Sample Input 2 :
2
20 16 -1 12 -1 8 -1 4 -1 -1 -1 
12
15 10 20 8 12 16 25 -1 -1 -1 -1 -1 -1 -1 -1
57
Sample Output 2:
4 8
-1 -1
Hint

Try to think of storing the inorder traversal in an array then try to find the nodes having sum equal to ‘S’.

Approaches (2)
Inorder Traversal

 The idea is to convert this problem to sorted array problem by storing the inorder traversal of BST in an array then finding the nodes having sum equal to ‘S’.

 

Approach :

  • First make a function ‘inorder’ passing the root and an array ‘arr’ to store the inorder traversal of BST as its arguments.
  • Inside ‘inorder()’ :
    • Store the inorder traversal of BST in ‘arr’.
  • Now, in the given function, call the inorder’ function to store the inorder traversal of BST in ‘arr’ and initialize a 2D array ‘ans’ to store the pair of nodes.
  • Initialise an iterator pointer ‘i’ with 0 and ‘j’ to n - 1, where n is the size of ‘arr’.
  • Iterate until i < j:
    • If arr[i] + arr[j] is equal to  ‘S’ then store the value of nodes in the ‘ans’ and then increment ‘i’ and decrement ‘j’.
    • Else if arr[i] + arr[j] > S then decrement ‘j’.
    • Else increment ‘i’.
  • Return ‘ans’ to the given function.
Time Complexity

O(N), Where ‘N’ denotes the number of nodes in BST.

 

As we are traversing all the nodes in the bst to check for the pairs having sum equal to ‘S’. Therefore, overall time complexity will be O(N).

Space Complexity

O(N), Where ‘N’ denotes the number of nodes in BST. 

 

As we are using auxiliary space to store all the nodes after inorder traversal. Therefore, overall space complexity will be O(N).

Code Solution
(100% EXP penalty)
Pair sum in a BST
Full screen
Console