Find Weighted Median Node in Tree

To solve this problem, efficiently traverse the tree using binary lifting and lowest common ancestor (LCA). By calculating the total path weight and tracking the edge weights dynamically, we can identify the median node on the path from each pair of query nodes. Understanding tree traversal and managing path sums is key to solving this problem.

Problem Statement

You are given an undirected, weighted tree with n nodes rooted at node 0. Each edge is represented by a 2D array where each element [ui, vi, wi] means an edge between nodes ui and vi with weight wi. For each query [uj, vj], determine the weighted median node on the path from node uj to node vj. The weighted median node is the first node on the path such that the sum of edge weights from uj to this node is greater than or equal to half the total path weight.

The problem requires efficiently processing multiple queries on a tree, where the size of the tree and the number of queries can be large. This is a graph traversal problem that involves calculating path weights dynamically, which can be achieved using techniques like binary lifting and lowest common ancestor (LCA) to optimize the solution for larger input sizes.

Examples

Example 1

Input: n = 2, edges = [[0,1,7]], queries = [[1,0],[0,1]]

Output: [0,1]

Example 2

Input: n = 3, edges = [[0,1,2],[2,0,4]], queries = [[0,1],[2,0],[1,2]]

Output: [1,0,2] E xplanation:

Example details omitted.

Example 3

Input: n = 5, edges = [[0,1,2],[0,2,5],[1,3,1],[2,4,3]], queries = [[3,4],[1,2]]

Output: [2,2]

Sum from 1 → 0 = 2 = 3.5 , median is node 2.

Constraints

• 2 <= n <= 105
• edges.length == n - 1
• edges[i] == [ui, vi, wi]
• 0 <= ui, vi < n
• 1 <= wi <= 109
• 1 <= queries.length <= 105
• queries[j] == [uj, vj]
• 0 <= uj, vj < n
• The input is generated such that edges represents a valid tree.

Solution Approach

Binary Lifting for Efficient Query Resolution

Binary lifting is used to preprocess the tree to allow for fast computation of the lowest common ancestor (LCA) of any two nodes. This allows for efficient path weight calculations by breaking the path into smaller, manageable segments, making it easier to find the weighted median node.

Path Sum Calculation

For each query, the path sum is calculated using the LCA to split the path into two segments. The sum of the edge weights is then dynamically tracked as the path is traversed from node uj to node vj, and the weighted median node is determined once the sum reaches half of the total path weight.

Handling Multiple Queries

To handle multiple queries efficiently, the tree is preprocessed using binary lifting and LCA, which allows for fast retrieval of path sums. This ensures that the solution can scale even for large input sizes, reducing the need for recalculating path sums for each query.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution depends on the preprocessing step for binary lifting, which takes O(n log n). Each query can then be answered in O(log n) time due to the efficient calculation of the LCA and path sum. Thus, for q queries, the overall time complexity is O(n log n + q log n). Space complexity is O(n log n) due to the storage required for binary lifting and path sum information.

What Interviewers Usually Probe

• Understanding binary lifting and LCA is crucial for an efficient solution.
• Ability to optimize graph traversal for multiple queries is important.
• The candidate should demonstrate a good grasp of dynamic state tracking during traversal.

Common Pitfalls or Variants

Common pitfalls

• Failing to optimize path sum calculations for multiple queries can lead to time inefficiencies.
• Incorrectly handling edge weights during path traversal, especially with large values.
• Not efficiently using binary lifting and LCA to reduce the time complexity for querying.

Follow-up variants

• Modify the problem to handle directed trees with varying edge directions.
• Add constraints that force the use of a specific traversal strategy (e.g., DFS or BFS).
• Extend the problem to handle dynamic edge weights that change during query resolution.

FAQ

How do I find the weighted median node in a tree?

The weighted median node is found by calculating the total path weight and tracking the sum of edge weights as you traverse the path from one node to another. The first node where the sum is greater than or equal to half of the total weight is the weighted median.

What is the role of binary lifting in this problem?

Binary lifting is used to preprocess the tree, enabling efficient calculation of the lowest common ancestor (LCA). This helps split the path into manageable parts, making it easier to calculate the weighted median for each query.

What are the time and space complexities of this problem?

The time complexity is O(n log n + q log n), where n is the number of nodes and q is the number of queries. Space complexity is O(n log n) due to the preprocessing for binary lifting and path sum information.

How can I optimize for multiple queries on the same tree?

By preprocessing the tree with binary lifting and LCA, you can efficiently answer multiple queries in O(log n) time, avoiding redundant path sum calculations.

What is the main difficulty of this problem?

The main difficulty lies in efficiently calculating path sums and finding the weighted median for multiple queries, especially when the tree is large and the number of queries is high.