Path 1: 15(1111)->3(0011)->5(0101) = 8
Path 2: 15(1111)->3(0011)->1(0001) = 7
Path 3: 15(01111)->7(00111)->31(11111) = 12 (maximum)
Path 4: 15(1111)->7(0111)->9(1001) = 9
Therefore, the maximum count of set bits obtained in a path is 12.
Path 1: 31(11111)->3(00011)->7(00111) = 10
Path 2: 31(11111)->3(00011)->1(00001) = 8
Path 3: 31(11111)->15(01111)->5(00101) = 11
Path 4: 31(11111)->15(01111)->23(10111) = 13 (maximum)
Therefore, the maximum count of set bits obtained in a path is 13.
Follow the steps below to solve the problem:
- Traverse each node recursively, starting from the root node
- Calculate the number of set bits in the value of the current node.
- Update the maximum count of set bits(stored in a variable, say maxm).
- Traverse its left and right subtree.
- After complete traversal of all the nodes of the tree, print the final value of maxm as the answer.
Below is the implementation of the above approach:
Time Complexity: O(N), where N denotes the number of nodes.
Auxiliary Space: O(1)
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