Given a tree, return a vector, with each elements of the vector is the sum of the distances between the corresponding node and all other nodes.

Intuition

Intuitively, I want to use DFS + memorization to solve the problem, however, such method exceeds te time limit.

Then, I refer to some solutions, and solve the problem.

Approach

DP -> TLE

In the first, I am thinking that we can use dp[i][j] to represent the distance between the node i and the node j. Initially, dp are initialized as follows:

If there is an edge between i and j, then dp[i][j]=dp[j][i]=1.
If there is an edge between i and j, then dp[i][j]=dp[j][i]=INFINITY.

we traversal from node 0 to node n-1, for each node i, we compute the distance between i and all other nodes j. There are two cases:

dp[i][j] != INFINITY, then we directly returns dp[i][j]
dp[i][j] == INFINITY, then it means the distance between node i and node j hasn’t been computed, we then update it as follows:

$$ dp[i][j] = \min_{k\in N(i)} (1 + dp[k][j]) $$

where $N(i)$ is the nodes that adjacent to node i. The min operation is used here since the node k and the node j may not connected (without passing node i).

The code is given as follows. However, the time complexity is $O(n^2)$, which exceeds the time limit.

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class Solution {
public:
    int dfs(const vector<vector<int>>& adjs, vector<vector<int>> &dp, 
        vector<bool> &visited, int start, int end) {
        if (dp[start][end] != INT_MAX)    return dp[start][end];
        if (visited[start]) return INT_MAX;
        visited[start] = true;
        int distance = INT_MAX;
        for (int next_node: adjs[start]) {
            distance = min(distance, dfs(adjs, dp, visited, next_node, end));
        }
        visited[start] = false;
        if (distance != INT_MAX)    dp[start][end] = distance + 1;
        return dp[start][end];
    }

    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        vector<vector<int>> dp(n, vector<int>(n, INT_MAX));
        vector<vector<int>> adjs(n);
        // initialize the distance matrix
        for (const auto& edge: edges) {
            dp[edge[0]][edge[1]] = 1;
            dp[edge[1]][edge[0]] = 1;
            adjs[edge[0]].push_back(edge[1]);
            adjs[edge[1]].push_back(edge[0]);
        }

        vector<bool> visited(n, false);
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                dfs(adjs, dp, visited, i, j);
            }
            // the graph is undirected
            for (int j = 0; j < i; ++j) {
                dp[i][j] = dp[j][i];
            }
        }

        vector<int> result(n);
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (j == i) continue;
                result[i] += dp[i][j];
            }
        }
        return result;
    }
};

Tree + Traversal

The second way is not easy to figure out. We decompose it into two parts:

compute the distance between the root node and all other nodes.
Convert the root node from one to another and update the result.

Sum of distance from root node to all other nodes.

Consider one example tree with root set as 0:

We first define dist[i] as the distance of all child nodes of node i to node i. Then we have:

$dist[i] = 0$ if i is a leaf node.
$dist[i] = \sum_{j\in C(i)}dist[j] + |C(i)|$ if i is not a leaf node, where $C(i)$ is the offspring of node i

We can now compute the sum of distances between root 0 to all other nodes, which is result[0].

Now we need to compute all other results. Repeating the above process is too time consuming, we need to reduce the time complexity. We are seeking a way to compute result[i] from result[j], where $j\in C(i)$.

Note that for $k\in C(j)$, when we compute result[i], we are computing distance from $k$ to $i$, so we need to reduce by 1 since their height decreases (the root changes from i to j). On the other hand, all other nodes, which are not offspring of node j, is added by 1 since the height of them increases. Thus, the transformation reads:

$$ result[j] = result[i] - |C(j)| + n - |C(j)| $$

Complexity

Time complexity: visited all nodes twice.
$$ O(n) $$
Space complexity: use two vectors of size $n$ to store the result.
$$ O(n) $$

Code

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class Solution {
public:
    void post_order_traversal(const vector<vector<int>>& adjs,
                              vector<int>& count, vector<int>& result,
                              int node, int parent) {
        for (int next_node : adjs[node]) {
            if (next_node == parent)
                continue;
            post_order_traversal(adjs, count, result, next_node, node);
            count[node] += count[next_node];
            result[node] += count[next_node] + result[next_node];
        }
        ++count[node];
    }

    void pre_order_traversal(const vector<vector<int>>& adjs,
                             vector<int>& count, vector<int>& result,
                             int node, int parent) {
        for (int next_node : adjs[node]) {
            if (next_node == parent)
                continue;
            result[next_node] =
                result[node] - count[next_node] + count.size() - count[next_node];
            pre_order_traversal(adjs, count, result, next_node, node);
        }
    }

    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        vector<int> count(n), result(n);
        vector<vector<int>> adjs(n);
        for (const auto &edge: edges) {
            adjs[edge[0]].push_back(edge[1]);
            adjs[edge[1]].push_back(edge[0]);
        }
        // compute result[0]
        post_order_traversal(adjs, count, result, 0, -1);
        // compute other results from result[0]
        pre_order_traversal(adjs, count, result, 0, -1);

        return result;
    }
};

834. Sum of Distances in Tree

Sum os distance of all nodes to all other nodes in a tree