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 when concatenated with the strings of the sub-tree nodes becomes a pangram.
Pangram: A pangram is a sentence containing every letter of the English Alphabet.
Only the weighted string of sub-tree of node 1 makes the pangram.
Approach: Perform dfs on the tree and update the weight of every node such that it stores its weight concatenated with the weights of the sub-tree nodes. Then, count the nodes whose updated weighted string forms a pangram.
Below is the implementation of the above approach:
- Subtree of all nodes in a tree using DFS
- Number of leaf nodes in the subtree of every node of an n-ary tree
- Convert a Binary Tree such that every node stores the sum of all nodes in its right subtree
- Change a Binary Tree so that every node stores sum of all nodes in left subtree
- Count the nodes in the given tree whose weight is even
- 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 given tree whose weight has X as a factor
- Count the nodes in the given tree whose weight is prime
- 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 nodes in the given tree whose weight is even parity
- 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
- Check if two nodes are in same subtree of the root node
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