Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is a power of 2.
Only the weight of the node 4 is a power of 2.
Approach: Perform dfs on the tree and for every node, check if its weight is a power of 2 or not, if yes then increment the count.
Below is the implementation of the above approach:
- Time Complexity: O(N).
In DFS, every node of the tree is processed once and hence the complexity due to the DFS is O(N) for N nodes in the tree. Therefore, the time complexity is O(N).
- Auxiliary Space: O(1).
Any extra space is not required, so the space complexity is constant.
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- Count of all prime weight nodes between given nodes in the given Tree
- Count the nodes in the given tree whose weight is even
- 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 in the given tree whose weight is even parity
- Count the nodes of the given tree whose weight has X as a factor
- Count of Nodes which has Prime Digit sum weight in a Tree
- Count the nodes in the given tree whose weight is a powerful number
- Count nodes in the given tree whose weight is a fibonacci number
- Count the nodes in the given Tree whose weight is a Perfect Number
- Query to find the maximum and minimum weight between two nodes in the given tree using LCA.
- 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 a given level in a tree using DFS
- 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 given level in a tree using BFS.
- Count the nodes of the tree whose weighted string contains a vowel
- Count of nodes in a Binary tree with immediate children as its factors
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