Diagonal Sum in Binary Tree (Algorithm)

Print the sum of every diagonal in binary tree.
Print diagonal elements: • Print diagonal element...

[@anandkulkarni2111]

Try to add some animation to your videos, You approach to explanation and step wise derivation is one of the best on you tube for algorithm related problems.!!!

[@vivekanandkhyade]

yess.... do u know someone who can help in making animations.. ?

[@pradumanyadav3828]

null point exception

[@pradumanyadav3828]

class Tree { public static ArrayList diagonalSum(Node root) { ArrayList List = new ArrayList(); path(root,List); return List; } public static void path(Node node,ArrayList lst){ Queue main = new ArrayDeque(); main.add(node); main.add(null); int sum = 0; while(main.size() > 0){ Node p = main.remove(); if(p == null){ lst.add(sum); main.add(null); sum = 0; p = main.remove(); if(p == null){ break; } } while(p != null){ sum += p.data; if(p.left != null){ main.add(p.left); } p = p.right; } } } }

[@pradumanyadav3828]

null point exception

[@saketp18]

I guess there is another way to do. Apply Hashmap where keys as distances and values as node data. Then add values and print the sum respective to key.

[@gautamjangir8927]

*C++ Solution* vector diagonalSum(Node* root) { vectorans; queueq; q.push(root); q.push(NULL); int sum = 0; while(!q.empty()) { Node* temp = q.front(); q.pop(); if(temp==NULL) { q.push(NULL); temp = q.front(); q.pop(); if(temp==NULL) { break; } } while(temp!=NULL) { sum=sum+temp->data; if(temp->left) { q.push(temp->left); } temp = temp->right; } if(q.front()==NULL)//This is the small change that requires in the code for proper execution { ans.push_back(sum); sum = 0; } } return ans; }

[@TechnicalBaniya]

I thing sir sum=0; statement should be in the the termination condition of the while loop because when p=dequeue(); statement will execute and then it is not NULL then sum will be initialized to zero.

[@vivekanandkhyade]

inside the inner while , that is added again...so we don't miss that........but your observation is really good...Thanks.

[@santhoshcts5]

I see another issue , when it dequeues and it receives null , it breaks the while and makes sum = 0 , even though the diagonal is not fully completed . take example of 4,9 when 9 has both right and left as empty , it breaks the inner while loop and goes to outer while loop and gets 10 , but during this time , sum is assigned as 0 and the value will be lost, even though there are still few elements in diagonal to complete adding to the sum . the correct way should be initialize sum as 0 only after printing the sum ( which means end of current diagonal and moving to next diagonal )

[@samcooper-02]

@@santhoshcts5 i agree

[@shivsagarshah24]

@@santhoshcts5 correct

[@shivsagarshah24]

@@vivekanandkhyade it should be written(sum = 0) after you have printed sum.

[@manassharma7192]

Sir ,sum=0 should be in the if condition after printing the sum.

[@shivsagarshah24]

Yes, you are right!!!

[@adityaranjan4639]

sir, your algorithm is good but there is some mistake at condition of sum where you initialize sum ,so it isn't giving proper output. please check once again. if we initialize sum=0 at declaration time and in the condition where we print sum. , my code is : if(p==NULL) { cout

[@rockyraj9571]

awesome... sri ji

[@nancygoel3473]

this code needs to be optimized

[@the-gray-haired-developer]

Appreciate your teaching techniques. For diagonal distance 3, you missed 15.

[@ankushroy919]

is it necessary to put null in queue we just check if root.left is null then push and traverse right

[@anshulasati1365]

if we dqueue 6 then sum is 0,so its calculate initially..please describe

[@DhananjayTyagi24]

Amazing explanation sirji

[@veerrajuyeleti8541]

sir can you upload a video on find the node in a binary tree having equal subtrees on right and left

[@veerrajuyeleti8541]

sir cAN WE INCLUDE HASH TABLE FOR STORING THE SAME DIAGONAL VALUES

[@vivekanandkhyade]

yes you can.

[@veerrajuyeleti8541]

well understood