Final answer:
To find the furthest node from each node in a tree using dynamic programming, implement two DFS passes to compute DP states for each node, considering both the children and parent contributions. The recursive approach has a base case for leaf nodes and builds up the solution for ancestor nodes, with a linear runtime complexity of O(n).
Step-by-step explanation:
To design a dynamic programming algorithm to find the furthest node from each node v within a tree T, we can use two passes of depth-first search (DFS). First, we need to find the longest path in the tree. Then, for each node, we can find the furthest node by combining the results from the children nodes and possibly the parent node.
The DP states can be defined as follows:
- dp[v][0] = the distance to the furthest node in the subtree rooted at v.
- dp[v][1] = the distance to the furthest node outside of the subtree rooted at v.
The base case would be for a leaf node where dp[v][0] = 0, as there are no nodes further within its subtree.
The recursive case would involve
- For each child u of a node v, calculate dp[u][0].
- Then, for each child u of v, consider the furthest distance from all other children subtrees and the parent node's calculated distance to find dp[u][1].
The correctness of this algorithm is justified by the fact that trees do not contain cycles, so the furthest distances can be computed directly based on the longest paths using the DP states. The runtime analysis of this algorithm would be O(n), where n is the number of nodes in the tree, since each node is visited exactly once in the two DFS passes.