Skip to main content

Diagonal Sum In Binary Tree| diagonal sum in Binary Tree Problem Solution

 diagonal sum in Binary Tree Problem Solution -

In this blog we will see How to print Diagonal Sum In Binary Tree| diagonal sum in Binary Tree Problem Solution . It is most common question related to recursion lets read diagonal sum in Binary Tree Problem.



Problem Statement-

Consider Red lines of slope -1 passing between nodes (in following diagram). The diagonal sum in a binary tree is the sum of all node’s data lying between these lines. Given a Binary Tree of size N, print all diagonal sums.

For the following input tree, output should be 9, 19, 42.
9 is sum of 1, 3 and 5.
19 is sum of 2, 6, 4 and 7.
42 is sum of 9, 10, 11 and 12.

DiagonalSum

Example 1:

Input:
         4
       /   \
      1     3
           /
          3
Output: 7 4 

Example 2:

Input:
           10
         /    \
        8      2
       / \    /
      3   5  2
Output: 12 15 3 

Your Task:
You don't need to take input. Just complete the function diagonalSum() that takes root node of the tree as parameter and returns an array containing the diagonal sums for every diagonal present in the tree with slope -1.

Expected Time Complexity: O(N).
Expected Auxiliary Space: O(N).

Constraints:
1<=Number of nodes<=105


Note: The Input/Ouput format and Example given are used for system's internal purpose, and should be used by a user for Expected Output only. As it is a function problem, hence a user should not read any input from stdin/console. The task is to complete the function specified, and not to write the full code.


Problem Credit - GeeksForGeeks (This Problem is taken From geeks for geeks website).


Recommended: Please try your approach before moving on to the solution.



Let's Jump To my Solution now-

After a deep observation, I divided this binary tree having n diagonal and I will write a function that will calculate all diagonal sum and store solution in a map.
for example, diagonal number 1 ans will be stored in a map having key 1.
see the image below




Fucntion Descreption ->

I have written a function that returns the sum of diagonal if I pass the current root, map(for storing diagonal sum till now ), and diagonal number to this function parameter. This is a magical function, isn't it 😆😆  XD. 

Now I have to find the smallest possible case which will help me to write base case of my problem 
if the root doesn't exist I have to simply return because the tree is empty.
Now I found the smallest problem  Lets write base case.

 if(!root){
        return ;
    }

Now How to write the next block of code.

Think like that -

I don't know the solution but if somehow i get diagonal sum excluded root node my task  will be minimized and i will  add only root node to previous sum.


`At the same time, I noticed that whenever I go to the left of my root node every time my diagonal number increased by 1. By Using this Property I can write my diagonal sum in Binary Tree Problem Solution.
so Let's write the induction step now if you think like that it will very easy to write induction step-


calculateDiagonalSum(diagonalNumber+1,root->left,mymap);
     calculateDiagonalSum(diagonalNumber,root->right,mymap);
     mymap[diagonalNumber]+=root->data;
     return ;

    Hopefully, You understood the entire solution if not comment below.
Here is the complete code of my solution -



Comments

Popular posts from this blog

Vaccine Distribution Codechef December Long challenge solution 2020

 Vaccine Distribution  Codechef December Long challenge solution 2020- This Problem vaccine distribution is taken from December long challenge 2020. lets read problem statement. Problem statement- Finally, a COVID vaccine is out on the market and the Chefland government has asked you to form a plan to distribute it to the public as soon as possible. There are a total of  N N  people with ages  a 1 , a 2 , … , a N a 1 , a 2 , … , a N . There is only one hospital where vaccination is done and it is only possible to vaccinate up to  D D  people per day. Anyone whose age is  ≥ 80 ≥ 80  or  ≤ 9 ≤ 9  is considered to be  at risk . On each day, you may not vaccinate both a person who is at risk and a person who is not at risk. Find the smallest number of days needed to vaccinate everyone. Input The first line of the input contains a single integer  T T  denoting the number of test cases. The description of  T T  test cases follows. The first line of each test case contains two space-separ

Codechef February long challenge solution 2021

   Codechef February  long challenge solution 2021- Hey guys Welcome Again, A small update CodeChef February  long challenge solution 2021 starting from 1st week of beb  2021 . Codechef long challenge is one of the popular contests of CodeChef you should participate in codechef Feb  long challenge.This is first contest of codechef in 2021 so good luck. And wish you happy new year. Contest link- https://www.codechef.com/FEB21/?itm_medium=navmenu&itm_campaign=FEB21 I will post the solution after Contest.Stay tuned. About CodeChef  February  long challenge solution 2021: CodeChef Long Challenge is a 10-day monthly coding contest where you can show off your computer programming skills. The significance being - it gives you enough time to think about a problem, try different ways of attacking the problem, read the concepts etc. If you’re usually slow at solving problems and have ample time at hand, this is ideal for you. We also put in a lot of efforts in getting quality problems, w

Merge Two sorted array Without Extra Space

Problem statement-  Given two sorted arrays arr1[] and arr2[] of sizes N and M in non-decreasing order. Merge them in sorted order without using any extra space. Modify arr1 so that it contains the first N elements and modify arr2 so that it contains the last M elements.    Example 1: Input:  N = 4, arr1[] = [1 3 5 7]  M = 5, arr2[] = [0 2 6 8 9] Output:  arr1[] = [0 1 2 3] arr2[] = [5 6 7 8 9] Explanation: After merging the two  non-decreasing arrays, we get,  0 1 2 3 5 6 7 8 9.   Example 2: Input:  N = 2, arr1[] = [10, 12]  M = 3, arr2[] = [5 18 20] Output:  arr1[] = [5 10] arr2[] = [12 18 20] Explanation: After merging two sorted arrays  we get 5 10 12 18 20. Your Task: You don't need to read input or print anything. You only need to complete the function merge() that takes arr1, arr2, N and M as input parameters and modifies them in-place so that they look like the sorted merged array when concatenated. Expected Time Complexity:  O((n+m) log(n+m)) Expected Auxilliary Space: O(1