Leetcode #834: Sum of Distances in Tree
In this guide, we solve Leetcode #834 Sum of Distances in Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There is an undirected connected tree with n nodes labeled from 0 to n - 1 and n - 1 edges. You are given the integer n and the array edges where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Tree, Depth-First Search, Graph, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: n = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]
Output: [8,12,6,10,10,10]
Explanation: The tree is shown above.
We can see that dist(0,1) + dist(0,2) + dist(0,3) + dist(0,4) + dist(0,5)
equals 1 + 1 + 2 + 2 + 2 = 8.
Hence, answer[0] = 8, and so on.
Python Solution
class Solution:
def sumOfDistancesInTree(self, n: int, edges: List[List[int]]) -> List[int]:
def dfs1(i: int, fa: int, d: int):
ans[0] += d
size[i] = 1
for j in g[i]:
if j != fa:
dfs1(j, i, d + 1)
size[i] += size[j]
def dfs2(i: int, fa: int, t: int):
ans[i] = t
for j in g[i]:
if j != fa:
dfs2(j, i, t - size[j] + n - size[j])
g = defaultdict(list)
for a, b in edges:
g[a].append(b)
g[b].append(a)
ans = [0] * n
size = [0] * n
dfs1(0, -1, 0)
dfs2(0, -1, ans[0])
return ans
Complexity
The time complexity is , and the space complexity is , where is the number of nodes in the tree. The space complexity is , where is the number of nodes in the tree.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.