Number of Ways to Assign Edge Weights I

This problem requires assigning each tree edge a weight of 1 or 2 such that path costs follow constraints. Use depth-first search to propagate valid states from the root to leaves, tracking cumulative weights carefully. By systematically exploring combinations, you can count all valid assignments efficiently without revisiting nodes.

Problem Statement

You are given an undirected tree with n nodes labeled from 1 to n, rooted at node 1. The tree is described by a 2D array edges of length n - 1, where edges[i] = [ui, vi] represents an edge between nodes ui and vi. Initially, all edges have weight 0. Assign each edge a weight of either 1 or 2 so that the total cost of any path satisfies implicit constraints derived from the tree structure.

The cost of a path between any two nodes u and v is the sum of weights along the unique path connecting them. Return the total number of valid assignments for all edges that maintain consistent path costs. Apply binary-tree traversal and depth-first search to track cumulative weights from the root and ensure all assignments follow the tree state rules.

Examples

Example 1

Input: edges = [[1,2]]

Output: 1

Example 2

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

Output: 2

Constraints

• 2 <= n <= 105
• edges.length == n - 1
• edges[i] == [ui, vi]
• 1 <= ui, vi <= n
• edges represents a valid tree.

Solution Approach

Depth-First Search for State Propagation

Traverse the tree starting from the root using DFS. At each node, maintain state information about cumulative path costs from the root. This allows you to check if assigning weight 1 or 2 to a child edge preserves valid path totals.

Binary-Tree Traversal and Assignment Counting

For each edge, consider both weight assignments and propagate the updated state to child nodes. Accumulate the number of valid assignments recursively. Avoid revisiting nodes by marking visited nodes to prevent double counting.

Modulo Handling and Final Aggregation

Because the number of valid assignments can grow exponentially, apply modulo operations during counting. After processing all edges via DFS, aggregate results from leaf to root to obtain the final number of valid edge weight assignments.

Complexity Analysis

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

Time complexity depends on DFS traversal of the tree, which is O(n) per path combination propagation. Space complexity is O(n) to store adjacency lists and state arrays for each node during traversal.

What Interviewers Usually Probe

• Pay attention to propagating cumulative path costs correctly during DFS.
• Check for overcounting by ensuring each node's state is updated only once per traversal.
• Consider using recursion with memoization to track edge weight assignments efficiently.

Common Pitfalls or Variants

Common pitfalls

• Assigning weights without updating cumulative path costs can lead to invalid totals.
• Failing to account for tree branching correctly results in missed or double-counted assignments.
• Ignoring modulo arithmetic may cause integer overflow in large trees.

Follow-up variants

• Changing edge weight options to {1,2,3} increases the state space and requires adjusted DFS counting.
• Calculating only paths between specific node pairs rather than all pairs simplifies cumulative state tracking.
• Handling rooted trees with additional constraints on leaf-to-root paths alters DFS propagation logic.

FAQ

What is the main idea behind Number of Ways to Assign Edge Weights I?

Use DFS to traverse the tree while tracking cumulative path costs from the root. Count valid assignments of 1 or 2 to each edge based on these states.

Can I use BFS instead of DFS for this problem?

BFS can work but DFS naturally handles recursive state propagation from root to leaves, making assignment counting simpler.

Why do we need to track cumulative weights at each node?

Cumulative weights ensure that all paths maintain consistent costs and prevent assigning edge weights that violate constraints.

How does modulo operation help in counting assignments?

It prevents integer overflow when the number of valid assignments becomes large, keeping computations manageable.

What is a common failure mode for this binary-tree traversal problem?

Failing to mark visited nodes or improperly propagating states can lead to double-counting or missed assignments in the DFS.