{ "data": { "question": { "questionId": "3842", "questionFrontendId": "3559", "categoryTitle": "Algorithms", "boundTopicId": 3683573, "title": "Number of Ways to Assign Edge Weights II", "titleSlug": "number-of-ways-to-assign-edge-weights-ii", "content": "
There is an undirected tree with n
nodes labeled from 1 to n
, rooted at node 1. The tree is represented by a 2D integer array edges
of length n - 1
, where edges[i] = [ui, vi]
indicates that there is an edge between nodes ui
and vi
.
Initially, all edges have a weight of 0. You must assign each edge a weight of either 1 or 2.
\n\nThe cost of a path between any two nodes u
and v
is the total weight of all edges in the path connecting them.
You are given a 2D integer array queries
. For each queries[i] = [ui, vi]
, determine the number of ways to assign weights to edges in the path such that the cost of the path between ui
and vi
is odd.
Return an array answer
, where answer[i]
is the number of valid assignments for queries[i]
.
Since the answer may be large, apply modulo 109 + 7
to each answer[i]
.
Note: For each query, disregard all edges not in the path between node ui
and vi
.
\n
Example 1:
\n\nInput: edges = [[1,2]], queries = [[1,1],[1,2]]
\n\nOutput: [0,1]
\n\nExplanation:
\n\n[1,1]
: The path from Node 1 to itself consists of no edges, so the cost is 0. Thus, the number of valid assignments is 0.[1,2]
: The path from Node 1 to Node 2 consists of one edge (1 → 2
). Assigning weight 1 makes the cost odd, while 2 makes it even. Thus, the number of valid assignments is 1.Example 2:
\n\nInput: edges = [[1,2],[1,3],[3,4],[3,5]], queries = [[1,4],[3,4],[2,5]]
\n\nOutput: [2,1,4]
\n\nExplanation:
\n\n[1,4]
: The path from Node 1 to Node 4 consists of two edges (1 → 3
and 3 → 4
). Assigning weights (1,2) or (2,1) results in an odd cost. Thus, the number of valid assignments is 2.[3,4]
: The path from Node 3 to Node 4 consists of one edge (3 → 4
). Assigning weight 1 makes the cost odd, while 2 makes it even. Thus, the number of valid assignments is 1.[2,5]
: The path from Node 2 to Node 5 consists of three edges (2 → 1, 1 → 3
, and 3 → 5
). Assigning (1,2,2), (2,1,2), (2,2,1), or (1,1,1) makes the cost odd. Thus, the number of valid assignments is 4.\n
Constraints:
\n\n2 <= n <= 105
edges.length == n - 1
edges[i] == [ui, vi]
1 <= queries.length <= 105
queries[i] == [ui, vi]
1 <= ui, vi <= n
edges
represents a valid tree.给你一棵有 n
个节点的无向树,节点从 1 到 n
编号,树以节点 1 为根。树由一个长度为 n - 1
的二维整数数组 edges
表示,其中 edges[i] = [ui, vi]
表示在节点 ui
和 vi
之间有一条边。
一开始,所有边的权重为 0。你可以将每条边的权重设为 1 或 2。
\n\n两个节点 u
和 v
之间路径的 代价 是连接它们路径上所有边的权重之和。
给定一个二维整数数组 queries
。对于每个 queries[i] = [ui, vi]
,计算从节点 ui
到 vi
的路径中,使得路径代价为 奇数 的权重分配方式数量。
返回一个数组 answer
,其中 answer[i]
表示第 i
个查询的合法赋值方式数量。
由于答案可能很大,请对每个 answer[i]
取模 109 + 7
。
注意: 对于每个查询,仅考虑 ui
到 vi
路径上的边,忽略其他边。
\n\n
示例 1:
\n\n输入: edges = [[1,2]], queries = [[1,1],[1,2]]
\n\n输出: [0,1]
\n\n解释:
\n\n[1,1]
:节点 1 到自身没有边,代价为 0,因此合法赋值方式为 0。[1,2]
:从节点 1 到节点 2 的路径有一条边(1 → 2
)。将权重设为 1 时代价为奇数,设为 2 时为偶数,因此合法赋值方式为 1。示例 2:
\n\n输入: edges = [[1,2],[1,3],[3,4],[3,5]], queries = [[1,4],[3,4],[2,5]]
\n\n输出: [2,1,4]
\n\n解释:
\n\n[1,4]
:路径为两条边(1 → 3
和 3 → 4
),(1,2) 或 (2,1) 的组合会使代价为奇数,共 2 种。[3,4]
:路径为一条边(3 → 4
),仅权重为 1 时代价为奇数,共 1 种。[2,5]
:路径为三条边(2 → 1 → 3 → 5
),组合 (1,2,2)、(2,1,2)、(2,2,1)、(1,1,1) 均为奇数代价,共 4 种。\n\n
提示:
\n\n2 <= n <= 105
edges.length == n - 1
edges[i] == [ui, vi]
1 <= queries.length <= 105
queries[i] == [ui, vi]
1 <= ui, vi <= n
edges
表示一棵合法的树。chainLength
and sumParity
.",
"Use Lowest Common Ancestor to find the distance between any two nodes quickly in O(logn)
."
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