Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is a Perfect number.
There is no node with a weight that is a perfect number.
In order to solve this problem, we perform Depth First Search(DFS) Traversal on the tree and for every node, check if its weight is a Perfect Number or not. We keep on incrementing the counter every time such a weight is obtained. The final value of that counter after the completion of the entire tree traversal is the answer.
Below is the implementation of the above approach:
- Count nodes in the given tree whose weight is a fibonacci number
- Count the nodes in the given tree whose weight is a powerful number
- Count the nodes whose weight is a perfect square
- Count of all prime weight nodes between given nodes in the given Tree
- Count the nodes in the given tree whose weight is even
- Count the nodes in the given tree whose sum of digits of weight is odd
- Count the nodes in the given tree whose weight is prime
- Count the nodes in the given tree whose weight is a power of two
- Count the nodes in the given tree whose weight is even parity
- Count the nodes of the given tree whose weight has X as a factor
- Count of Nodes which has Prime Digit sum weight in a Tree
- Count number of paths whose weight is exactly X and has at-least one edge of weight M
- Construct XOR tree by Given leaf nodes of Perfect Binary Tree
- Query to find the maximum and minimum weight between two nodes in the given tree using LCA.
- Count the number of nodes at given level in a tree using BFS.
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Improved By : princi singh