Given a tree, and the weights (in the form of strings) of all the nodes, the task is to count the nodes whose weighted string is an anagram with the given string str.
str = “geek”
Only the weighted strings of the nodes 2 and 6
are anagrams of the given string “geek”.
Approach: Perform dfs on the tree and for every node, check if it’s weighted string is anagram with the given string or not, If not then increment the count.
Below is the implementation of the above approach:
- 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
- Count the nodes of a tree whose weighted string does not contain any duplicate characters
- Sum of nodes at k-th level in a tree represented as string
- Product of nodes at k-th level in a tree represented as string
- Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes
- 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 Non-Leaf nodes in a Binary Tree
- Determine the count of Leaf nodes in an N-ary tree
- Count the number of nodes at given level in a tree using BFS.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.