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Problem solution · Python

Shortest Path in a Weighted Tree

Shortest Path in a Weighted Tree: a Python solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Segment tree or range structure
Source
Kamyu LeetCode Solutions
Length
130 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Segment tree or range structure

For Shortest Path in a Weighted Tree, the implementation stores interval information in a range-query data structure so updates and queries avoid rescanning the full input.

  1. Choose the aggregate stored for each interval or prefix.
  2. Build or initialize the structure from the input.
  3. Apply updates and combine the affected nodes to answer each query.

Code notes

  • 130 lines of Python from the credited upstream file shortest-path-in-a-weighted-tree.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count the build once, then multiply the logarithmic update or query path by the number of operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeShortest Path in a Weighted Tree · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(nlogn)# Space: O(n) class BIT(object):  # 0-indexed.    def __init__(self, n):        self.__bit = [0]*(n+1)  # Extra one for dummy node.     def add(self, i, val):        i += 1  # Extra one for dummy node.        while i < len(self.__bit):            self.__bit[i] += val            i += (i & -i)     def query(self, i):        i += 1  # Extra one for dummy node.        ret = 0        while i > 0:            ret += self.__bit[i]            i -= (i & -i)        return ret  # iterative dfs, fenwick treeclass Solution(object):    def treeQueries(self, n, edges, queries):        """        :type n: int        :type edges: List[List[int]]        :type queries: List[List[int]]        :rtype: List[int]        """        def iter_dfs():            L, R, dist, lookup = [0]*n, [0]*n, [0]*n, [0]*n            cnt = 0            stk = [(1, (0, -1, 0))]            while stk:                step, args = stk.pop()                if step == 1:                    u, p, d = args                    L[u] = cnt                    cnt += 1                    dist[u] = d                    stk.append((2, (u,)))                    for v, w in adj[u]:                        if v == p:                            continue                        lookup[v] = w                        stk.append((1, (v, u, d+w)))                elif step == 2:                    u = args[0]                    R[u] = cnt            return L, R, dist, lookup         adj = [[] for _ in xrange(n)]        for u, v, w in edges:            u -= 1            v -= 1            adj[u].append((v, w))            adj[v].append((u, w))        L, R, dist, lookup = iter_dfs()        bit = BIT(n)        result = []        for q in queries:            if q[0] == 1:                _, u, v, w = q                u -= 1                v -= 1                if L[u] > L[v]:                    u, v = v, u                 diff = w-lookup[v]                bit.add(L[v], diff)                bit.add(R[v], -diff)                lookup[v] = w            else:                _, x = q                x -= 1                result.append(dist[x]+bit.query(L[x]))        return result  # Time:  O(nlogn)# Space: O(n)# dfs, fenwick treeclass Solution2(object):    def treeQueries(self, n, edges, queries):        """        :type n: int        :type edges: List[List[int]]        :type queries: List[List[int]]        :rtype: List[int]        """        def dfs(u, p, d):            L[u] = cnt[0]            cnt[0] += 1            dist[u] = d            for v, w in adj[u]:                if v == p:                    continue                lookup[v] = w                dfs(v, u, d+w)            R[u] = cnt[0]         adj = [[] for _ in xrange(n)]        for u, v, w in edges:            u -= 1            v -= 1            adj[u].append((v, w))            adj[v].append((u, w))        L, R, dist, lookup = [0]*n, [0]*n, [0]*n, [0]*n        cnt = [0]        dfs(0, -1, 0)        bit = BIT(n)        result = []        for q in queries:            if q[0] == 1:                _, u, v, w = q                u -= 1                v -= 1                if L[u] > L[v]:                    u, v = v, u                 diff = w-lookup[v]                bit.add(L[v], diff)                bit.add(R[v], -diff)                lookup[v] = w            else:                _, x = q                x -= 1                result.append(dist[x]+bit.query(L[x]))        return result

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