Given a tree and the weights of all the nodes, the task is to count the number of nodes whose weights are even parity i.e. whether the count of set bits in them is even.
Weight Binary Representation Parity 5 0101 Even 10 1010 Even 11 1011 Odd 8 1000 Odd 6 0110 Even
Approach: Perform dfs on the tree and for every node, check if it’s weight is even parity 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 even
- 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 of the given tree whose weight has X as a factor
- 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 the number of nodes at given level in a tree using BFS.
- Count the number of nodes at a given level in a tree using DFS
- Count Non-Leaf nodes in a Binary Tree
- Determine the count of Leaf nodes in an N-ary tree
- 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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