How do you solve Shortest Path in a Weighted Tree on LeetCode?

Shortest Path in a Weighted Tree (LeetCode #3515) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(n) space complexity. Key insight: Euler tour helps flatten the tree.

What is the time complexity of Shortest Path in a Weighted Tree?

The optimal solution for Shortest Path in a Weighted Tree runs in O(log n) time and uses O(n) space. Segment tree operations (update and query) are O(log n), making the overall complexity efficient for multiple queries.

What is the brute force solution for Shortest Path in a Weighted Tree?

The brute force approach for Shortest Path in a Weighted Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. For each query, traverse the tree from the root to the target node, summing the weights of edges. This approach is straightforward but inefficient for large trees. The optimal Optimal Solution approach improves this to O(log n).

What are the interview tips for Shortest Path in a Weighted Tree?

Explain your thought process clearly. Discuss trade-offs of different approaches. Practice coding segment trees.

What are common mistakes when solving Shortest Path in a Weighted Tree?

Not handling edge cases in updates. Forgetting to maintain tree structure during updates.

#3515

Shortest Path in a Weighted Tree

Hard

ArrayTreeDepth-First SearchBinary Indexed TreeSegment TreeTree TraversalSegment Tree

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n)
Space	O(1)	O(n)

💡

Intuition

Time O(log n)Space O(n)

Use an Euler tour to flatten the tree, allowing efficient updates and queries via a segment tree. This reduces the complexity of path calculations significantly.

⚙️

Algorithm

3 steps

1Step 1: Perform an Euler tour to create a flattened representation of the tree and record edge weights.
2Step 2: Build a segment tree over the flattened array to support efficient updates and queries.
3Step 3: For each query, either update the segment tree or retrieve the distance using point queries.

solution.py37 lines

1class SegmentTree:
2    def __init__(self, data):
3        self.n = len(data)
4        self.tree = [0] * (2 * self.n)
5        for i in range(self.n): self.tree[self.n + i] = data[i]
6        for i in range(self.n - 1, 0, -1): self.tree[i] = self.tree[i * 2] + self.tree[i * 2 + 1]
7    def update(self, idx, value):
8        idx += self.n
9        self.tree[idx] = value
10        while idx > 1:
11            idx //= 2
12            self.tree[idx] = self.tree[idx * 2] + self.tree[idx * 2 + 1]
13    def query(self, left, right):
14        res = 0
15        left += self.n
16        right += self.n
17        while left < right:
18            if left % 2:
19                res += self.tree[left]
20                left += 1
21            if right % 2:
22                right -= 1
23                res += self.tree[right]
24            left //= 2
25            right //= 2
26        return res
27
28def shortest_path_optimal(n, edges, queries):
29    # Build the tree and perform Euler tour (not implemented)
30    segment_tree = SegmentTree([])  # Placeholder for segment tree initialization
31    results = []
32    for query in queries:
33        if query[0] == 1:
34            segment_tree.update(...)  # Update logic
35        elif query[0] == 2:
36            results.append(segment_tree.query(...))  # Query logic
37    return results

ℹ

Complexity note: Segment tree operations (update and query) are O(log n), making the overall complexity efficient for multiple queries.

1Euler tour helps flatten the tree.
2Segment tree allows efficient updates and queries.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.