Given a tree, and the weights (in the form of strings) of all the nodes, the task is to count the nodes whose weights are palindrome.
Only the weights of the nodes 2, 3 and 5 are palindromes.
Approach: Perform dfs on the tree and for every node, check if it’s string is palindrome or not. If yes then increment the count.
Below is the implementation of the above approach:
- 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
- Count the nodes of a tree whose weighted string is an anagram of the given string
- Minimum distance to visit all the nodes of an undirected weighted tree
- Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes
- Count All Palindrome Sub-Strings in a String | Set 2
- 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 sum of digits of weight is odd
- Count the nodes of the given tree whose weight has X as a factor
- Count the number of nodes at a given level in a tree using DFS
- Count the nodes in the given tree whose weight is prime
- Determine the count of Leaf nodes in an N-ary tree
- Count the nodes in the given tree whose weight is even parity
- Count Non-Leaf nodes in a Binary Tree
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