Sum of Distances in Tree

Problem Statement

In an undirected connected tree with n nodes labeled from 0 to n - 1, connected by n - 1 edges, your task is to compute specific values for each node based on the node's connectivity and tree structure. Each tree edge connects two nodes ai and bi, as specified by an array edges where each element edges[i] is a pair [ai, bi].

The goal is to return an array answer where each item answer[i] contains the total sum of distances from the i-th node to every other node in the tree. In other words, for each node, calculate the sum of the shortest paths from this node to every other node and populate these sums in the resultant array.

Examples

Example 1

Input:

n = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]

Output:

[8,12,6,10,10,10]

Explanation:

The tree is shown above.
We can see that dist(0,1) + dist(0,2) + dist(0,3) + dist(0,4) + dist(0,5)
equals 1 + 1 + 2 + 2 + 2 = 8.
Hence, answer[0] = 8, and so on.

Example 2

Input:

n = 1, edges = []

Output:

[0]

Example 3

Input:

n = 2, edges = [[1,0]]

Output:

[1,1]

Constraints

• 1 <= n <= 3 * 104
• edges.length == n - 1
• edges[i].length == 2
• 0 <= ai, bi < n
• ai != bi
• The given input represents a valid tree.

Approach and Intuition

1. This problem requires an understanding of tree data structures, how to traverse them, and calculate distances between nodes in an efficient manner.
2. Given the constraints on node numbers (n <= 30,000), an efficient algorithm must avoid direct path calculation between all pairs, as that would be computationally expensive.
3. Efficient traversal methods such as Depth-First Search (DFS) can be employed to calculate distances from a root node to all other nodes initially.

Here’s how we can visualize and solve the problem:

• Initial Setup: Convert the edge list into an adjacency list for easy access during traversal.
• First Pass (Rooting the Tree):
  • Choose any node as the root. Often, node 0 is used as a convenient root.
  • Apply DFS/BFS from the root to calculate initial distances to all nodes and also count the number of nodes in each subtree rooted at every node.
• Second Pass (Using Parent Distances):
  • Use the distances computed from the root node to deduce distances for the rest of the nodes using the properties of the tree:
    • If you know the distance from a node A to all other nodes, and if you move to its child B, the only changes in distances are for the nodes in subtree rooted at B and all other nodes.
    • Specifically, moving from A to B, the nodes in the subtree of B get closer by 1, and all others get farther by 1.
  • This can help compute distances for each node based on already computed distances for its parent, significantly lowering the computational overhead.

Given examples show how the distances sum up for small-sized trees and how sum of distances remains constant when only two nodes are connected. Through this intuitive yet analytic approach, each node’s distances to others are effectively and efficiently computed, abiding by given constraints and expected tree characteristics.

Solutions

class TreeCalculator {
    int[] result, subtreeSize;
    List<Set<Integer>> treeStructure;
    int totalNodes;
    public int[] calculateDistanceSum(int totalNodes, int[][] edges) {
        this.totalNodes = totalNodes;
        treeStructure = new ArrayList<Set<Integer>>();
        result = new int[totalNodes];
        subtreeSize = new int[totalNodes];
        Arrays.fill(subtreeSize, 1);

        for (int i = 0; i < totalNodes; ++i)
            treeStructure.add(new HashSet<Integer>());
        for (int[] edge: edges) {
            treeStructure.get(edge[0]).add(edge[1]);
            treeStructure.get(edge[1]).add(edge[0]);
        }
        postOrderTraverse(0, -1);
        preOrderTraverse(0, -1);
        return result;
    }

    public void postOrderTraverse(int node, int parent) {
        for (int neighbor: treeStructure.get(node)) {
            if (neighbor != parent) {
                postOrderTraverse(neighbor, node);
                subtreeSize[node] += subtreeSize[neighbor];
                result[node] += result[neighbor] + subtreeSize[neighbor];
            }
        }
    }

    public void preOrderTraverse(int node, int parent) {
        for (int neighbor: treeStructure.get(node)) {
            if (neighbor != parent) {
                result[neighbor] = result[node] - subtreeSize[neighbor] + totalNodes - subtreeSize[neighbor];
                preOrderTraverse(neighbor, node);
            }
        }
    }
}