Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose sum of digits of weights is odd.
Node 1: digitSum(144) = 1 + 4 + 4 = 9
Node 2: digitSum(1234) = 1 + 2 + 3 + 4 = 10
Node 3: digitSum(21) = 2 + 1 = 3
Node 4: digitSum(5) = 5
Node 5: digitSum(77) = 7 + 7 = 14
Only the sum of digits of the weights of nodes 1, 3 and 4 are odd.
Approach: Perform dfs on the tree and for every node, check if the sum of the digits of its weight is odd. 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 of the given tree whose weight has X as a factor
- 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 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 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 Non-Leaf nodes in a Binary Tree
- Count the nodes of the tree whose weighted string contains a vowel
- 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
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