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The real breakthrough happens during the second run of DFS.

There is an undirected tree where each vertex is numbered from to, and each contains a data sum of a tree is the sum of all its nodes' data values. If an edge is cut, two smaller trees are formed. The difference between two trees is the absolute value of the difference in their sums. Given a tree, determine which edge to cut so that the resulting trees have a minimal. For those who are stuck, here are hints: 1. Dont think of it as a Tree problem. Approach it as a GRAPH. 2.

Use first run of DFS to compute a value (not telling what it is:p) that you think will help you solve the problem.

3. The real breakthrough happens during the second run of shrubdisposal.buzzted Reading Time: 6 mins. Jun 02, 1. Users who have contributed to this file. 54 lines (50 sloc) Bytes. Raw Blame. Open with Desktop. View raw. View blame. // //Cut the tree.

Answer (1 of 3): This problem is a simple application of Recursion and Depth First Traversal of a tree. You need to solve this problem using post order traversal technique, considering the tree as a rooted tree. (Hint: During dfs, each edge will be traversed only once, thats when you find T1. Solution of hackerrank problem solving. Contribute to Decent-Coder/Hackerrank-problem-solving development by creating an account on GitHub.