Given a tree, and the weights of all the nodes, the task is to count the nodes whose weights are divisible by x.
x = 5
Only the nodes 1 and 2 have weights divisible by 5.
Approach: Perform dfs on the tree and for every node, check if it’s weight is divisible by x or not. If yes then increment the count.
Below is the implementation of the above approach:
- 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 weight is prime
- Count the nodes in the given tree whose sum of digits of weight is odd
- Count the nodes whose weight is a perfect square
- Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes
- Count number of paths whose weight is exactly X and has at-least one edge of weight M
- Count Non-Leaf nodes in a Binary Tree
- Count the number of nodes at a given level in a tree using DFS
- Determine the count of Leaf nodes in an N-ary tree
- Count the number of nodes at given level in a tree using BFS.
- Count the nodes of the given tree whose weighted string is a palindrome
- Program to count leaf nodes in a binary tree
- Count nodes with two children at level L in a Binary Tree
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