Given a Binary Tree having positive and negative nodes, the task is to find the maximum sum level in it.
Input : 4 / \ 2 -5 / \ /\ -1 3 -2 6 Output: 6 Explanation : Sum of all nodes of 0'th level is 4 Sum of all nodes of 1'th level is -3 Sum of all nodes of 0'th level is 6 Hence maximum sum is 6 Input : 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 Output : 17
This problem is a variation of maximum width problem. The idea is to do level order traversal of tree. While doing traversal, process nodes of different level separately. For every level being processed, compute sum of nodes in the level and keep track of maximum sum.
Below is the implementation of the above idea:
Maximum level sum is 17
Time Complexity : O(N) where N is the total number of nodes in the tree.
In level order traversal, every node of the tree is processed once and hence the complexity due to the level order traversal is O(N) if there are total N nodes in the tree. Also, while processing every node, we are maintaining the sum at each level, however, this does not affect the overall time complexity. Therefore, the time complexity is O(N).
Auxiliary Space : O(w) where w is the maximum width of the tree.
In level order traversal, a queue is maintained whose maximum size at any moment can go upto the maximum width of the binary tree.
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Improved By : muskan_garg