Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is even.
Only the weights of the nodes 2, 4 and 5 are even.
Approach: Perform dfs on the tree and for every node, check if it’s weight is divisible by 2 or not. If yes then increment count.
Below is the implementation of the above approach:
- 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 in the given tree whose sum of digits of weight is odd
- Count the nodes of the given tree whose weight has X as a factor
- Count the nodes in the given tree whose weight is prime
- 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
- Determine the count of Leaf nodes in an N-ary tree
- Count Non-Leaf nodes in a Binary Tree
- Count the number of nodes at a given level in a tree using DFS
- Count the number of nodes at given level in a tree using BFS.
- Count nodes with two children at level L in a Binary Tree
- Count the nodes of the given tree whose weighted string is a palindrome
- Count the nodes of the tree whose weighted string contains a vowel
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