LeetCode 834: Sum of Distances in Tree Solution

Master LeetCode problem 834 (Sum of Distances in Tree), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

834. Sum of Distances in Tree

Hard

Dynamic Programming Tree Depth-First Search Graph

Problem Explanation

Explanation:

To solve this problem, we can use a depth-first search (DFS) approach. We need to calculate the sum of distances for each node in the tree. The idea is to calculate two values for each node:

count[node]: the number of nodes in the subtree rooted at node.
res[node]: the sum of distances from node to all other nodes.

We start by doing a post-order traversal to calculate count and res for each node. Then, using the values calculated during the post-order traversal, we do a pre-order traversal to update the res values for all nodes by propagating the information from their parent nodes. :

Solution Code

class Solution {
    int[] res, count;
    List<Set<Integer>> tree;
    
    public int[] sumOfDistancesInTree(int n, int[][] edges) {
        res = new int[n];
        count = new int[n];
        tree = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            tree.add(new HashSet<>());
        }
        
        for (int[] edge : edges) {
            tree.get(edge[0]).add(edge[1]);
            tree.get(edge[1]).add(edge[0]);
        }
        
        postOrder(0, -1);
        preOrder(0, -1);
        
        return res;
    }
    
    private void postOrder(int node, int parent) {
        for (int child : tree.get(node)) {
            if (child == parent) continue;
            postOrder(child, node);
            count[node] += count[child];
            res[node] += res[child] + count[child];
        }
        count[node]++;
    }
    
    private void preOrder(int node, int parent) {
        for (int child : tree.get(node)) {
            if (child == parent) continue;
            res[child] = res[node] - count[child] + count.length - count[child];
            preOrder(child, node);
        }
    }
}

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