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:
- 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 the number of nodes at given level in a tree using BFS.
- Count the nodes of the given tree whose weighted string is a palindrome
- Program to count leaf nodes in a binary tree
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