LeetCode #834 — HARD

Sum of Distances in Tree

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i in the tree.

Return an array answer of length n where answer[i] is the sum of the distances between the i^th node in the tree and all other nodes.

Example 1:

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.

Example 2:

Input: n = 1, edges = []
Output: [0]

Example 3:

Input: n = 2, edges = [[1,0]]
Output: [1,1]

Constraints:

1 <= n <= 3 * 10⁴
edges.length == n - 1
edges[i].length == 2
0 <= a_i, b_i < n
a_i != b_i
The given input represents a valid tree.

Step 01

Brute Force Baseline

Problem summary: 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. Return an array answer of length n where answer[i] is the sum of the distances between the ith node in the tree and all other nodes.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming · Tree

Example 1

6
[[0,1],[0,2],[2,3],[2,4],[2,5]]

Example 2

1
[]

Example 3

2
[[1,0]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #834: Sum of Distances in Tree
class Solution {
    private int n;
    private int[] ans;
    private int[] size;
    private List<Integer>[] g;

    public int[] sumOfDistancesInTree(int n, int[][] edges) {
        this.n = n;
        g = new List[n];
        ans = new int[n];
        size = new int[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        dfs1(0, -1, 0);
        dfs2(0, -1, ans[0]);
        return ans;
    }

    private void dfs1(int i, int fa, int d) {
        ans[0] += d;
        size[i] = 1;
        for (int j : g[i]) {
            if (j != fa) {
                dfs1(j, i, d + 1);
                size[i] += size[j];
            }
        }
    }

    private void dfs2(int i, int fa, int t) {
        ans[i] = t;
        for (int j : g[i]) {
            if (j != fa) {
                dfs2(j, i, t - size[j] + n - size[j]);
            }
        }
    }
}

// Accepted solution for LeetCode #834: Sum of Distances in Tree
func sumOfDistancesInTree(n int, edges [][]int) []int {
	g := make([][]int, n)
	for _, e := range edges {
		a, b := e[0], e[1]
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	ans := make([]int, n)
	size := make([]int, n)
	var dfs1 func(i, fa, d int)
	dfs1 = func(i, fa, d int) {
		ans[0] += d
		size[i] = 1
		for _, j := range g[i] {
			if j != fa {
				dfs1(j, i, d+1)
				size[i] += size[j]
			}
		}
	}
	var dfs2 func(i, fa, t int)
	dfs2 = func(i, fa, t int) {
		ans[i] = t
		for _, j := range g[i] {
			if j != fa {
				dfs2(j, i, t-size[j]+n-size[j])
			}
		}
	}
	dfs1(0, -1, 0)
	dfs2(0, -1, ans[0])
	return ans
}

# Accepted solution for LeetCode #834: Sum of Distances in Tree
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

// Accepted solution for LeetCode #834: Sum of Distances in Tree
/**
 * [0834] Sum of Distances in Tree
 *
 * 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.
 * Return an array answer of length n where answer[i] is the sum of the distances between the i^th node in the tree and all other nodes.
 *  
 * Example 1:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-sumdist1.jpg" style="width: 304px; height: 224px;" />
 * 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.
 *
 * Example 2:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-sumdist2.jpg" style="width: 64px; height: 65px;" />
 * Input: n = 1, edges = []
 * Output: [0]
 *
 * Example 3:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-sumdist3.jpg" style="width: 144px; height: 145px;" />
 * Input: n = 2, edges = [[1,0]]
 * Output: [1,1]
 *
 *  
 * Constraints:
 *
 * 	1 <= n <= 3 * 10^4
 * 	edges.length == n - 1
 * 	edges[i].length == 2
 * 	0 <= ai, bi < n
 * 	ai != bi
 * 	The given input represents a valid tree.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/sum-of-distances-in-tree/
// discuss: https://leetcode.com/problems/sum-of-distances-in-tree/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    // Credit: https://leetcode.com/problems/sum-of-distances-in-tree/discuss/892069/Rust-translated-8ms-100
    pub fn sum_of_distances_in_tree(n: i32, edges: Vec<Vec<i32>>) -> Vec<i32> {
        let mut tree = vec![std::collections::HashSet::<i32>::new(); n as usize];
        let mut result = vec![0; n as usize];
        let mut count = vec![0; n as usize];

        for edge in &edges {
            tree[edge[0] as usize].insert(edge[1]);
            tree[edge[1] as usize].insert(edge[0]);
        }

        Self::dfs(&tree, &mut count, &mut result, 0, -1);
        Self::dfs2(&tree, &mut count, &mut result, 0, -1);

        result
    }

    fn dfs(
        tree: &[std::collections::HashSet<i32>],
        count: &mut Vec<i32>,
        result: &mut Vec<i32>,
        root: i32,
        pre: i32,
    ) {
        for &i in &tree[root as usize] {
            if i == pre {
                continue;
            }
            Self::dfs(tree, count, result, i, root);
            count[root as usize] += count[i as usize];
            result[root as usize] += result[i as usize] + count[i as usize];
        }
        count[root as usize] += 1;
    }

    fn dfs2(
        tree: &[std::collections::HashSet<i32>],
        count: &mut Vec<i32>,
        result: &mut Vec<i32>,
        root: i32,
        pre: i32,
    ) {
        for &i in &tree[root as usize] {
            if i == pre {
                continue;
            }
            result[i as usize] =
                result[root as usize] - count[i as usize] + count.len() as i32 - count[i as usize];
            Self::dfs2(tree, count, result, i, root);
        }
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_0834_example_1() {
        let n = 6;
        let edges = vec![vec![0, 1], vec![0, 2], vec![2, 3], vec![2, 4], vec![2, 5]];
        let result = vec![8, 12, 6, 10, 10, 10];

        assert_eq!(Solution::sum_of_distances_in_tree(n, edges), result);
    }

    #[test]
    fn test_0834_example_2() {
        let n = 1;
        let edges: Vec<Vec<i32>> = vec![];
        let result = vec![0];

        assert_eq!(Solution::sum_of_distances_in_tree(n, edges), result);
    }

    #[test]
    fn test_0834_example_3() {
        let n = 2;
        let edges: Vec<Vec<i32>> = vec![vec![1, 0]];
        let result = vec![1, 1];

        assert_eq!(Solution::sum_of_distances_in_tree(n, edges), result);
    }
}

// Accepted solution for LeetCode #834: Sum of Distances in Tree
function sumOfDistancesInTree(n: number, edges: number[][]): number[] {
    const g: number[][] = Array.from({ length: n }, () => []);
    for (const [a, b] of edges) {
        g[a].push(b);
        g[b].push(a);
    }
    const ans: number[] = new Array(n).fill(0);
    const size: number[] = new Array(n).fill(0);
    const dfs1 = (i: number, fa: number, d: number) => {
        ans[0] += d;
        size[i] = 1;
        for (const j of g[i]) {
            if (j !== fa) {
                dfs1(j, i, d + 1);
                size[i] += size[j];
            }
        }
    };
    const dfs2 = (i: number, fa: number, t: number) => {
        ans[i] = t;
        for (const j of g[i]) {
            if (j !== fa) {
                dfs2(j, i, t - size[j] + n - size[j]);
            }
        }
    };
    dfs1(0, -1, 0);
    dfs2(0, -1, ans[0]);
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.