Given a Binary Tree, find vertical sum of the nodes that are in same vertical line. Print all sums through different vertical lines starting from left-most vertical line to right-most vertical line.
Example 1:
Input: 1
/ \
2 3
/ \ / \
4 5 6 7
Output:
Explanation:The tree has 5 vertical lines
Vertical-Line-1 has only one node
4 => vertical sum is 4
Vertical-Line-2: has only one node
2=> vertical sum is 2
Vertical-Line-3: has three nodes:
1,5,6 => vertical sum is 1+5+6 = 12
Vertical-Line-4: has only one node 3
=> vertical sum is 3
Vertical-Line-5: has only one node 7
=> vertical sum is 7
Your Task: You don't need to take input. Just complete the function verticalSum() that takes root node of the treeas parameter and returns an array containing the vertical sum of tree from left to right.
Expected Time Complexity: O(N). Expected Auxiliary Space: O(N).
Reset the IDE using the second button on the top right corner.
Avoid using static/global variables in your code as your code is tested
against multiple test cases and these tend to retain their previous values.
Passing the Sample/Custom Test cases does not guarantee the correctness of code.
On submission, your code is tested against multiple test cases consisting of all
possible corner cases and stress constraints.
You can access the hints to get an idea about what is expected of you as well as
the final solution code.
You can view the solutions submitted by other users from the submission tab.
Here is some troubleshooting advice for contest problems
Make sure you are not using ad-blockers.
Disable browser extensions.
We recommend using latest version of your browser for best experience.
Avoid using static/global variables in coding problems as your code is tested
against multiple test cases and these tend to retain their previous values.
Passing the Sample/Custom Test cases in coding problems does not guarantee the
correctness of code. On submission, your code is tested against multiple test cases
consisting of all possible corner cases and stress constraints.
POTD
POTD
