Given a tree with the weights of all the nodes, the task is to count the number of nodes whose weight is a Fibonacci number.
Nodes having weights 5 and 8 are fibonacci nodes.
Nodes having weights 1, 3 and 8 are fibonacci nodes.
Approach: The idea is to perform a dfs on the tree and for every node, check whether the weight is a Fibonacci number or not.
- Generate a hash containing all the Fibonacci numbers using Dynamic programming.
- Using depth-first search traversal, traverse through every node of the tree and check whether the node is a Fibonacci number or not by checking if that element is present in the precomputed hash or not.
- Finally, print the total number of Fibonacci nodes.
Below is the implementation of above approach:
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- Count the nodes in the given Tree whose weight is a Perfect Number
- Count the nodes in the given tree whose weight is a powerful number
- 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 weight is even parity
- Count the nodes in the given tree whose weight is a power of two
- 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 of the given tree whose weight has X as a factor
- Count of Nodes which has Prime Digit sum weight in a Tree
- Count the nodes whose sum with X is a Fibonacci number
- Count number of paths whose weight is exactly X and has at-least one edge of weight M
- Count the nodes whose weight is a perfect square
- Query to find the maximum and minimum weight between two nodes in the given tree using LCA.
- Count the number of nodes at a given level in a tree using DFS
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