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Problem solution · C++

Maximum Distinct Path Sum in a Binary Tree

Maximum Distinct Path Sum in a Binary Tree: a C++ solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Breadth-first search
Source
Kamyu LeetCode Solutions
Length
121 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

Breadth-first search

For Maximum Distinct Path Sum in a Binary Tree, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 121 lines of C++ from the credited upstream file maximum-distinct-path-sum-in-a-binary-tree.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 9 loop blocks detected, together with recursive traversal.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

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 codeMaximum Distinct Path Sum in a Binary Tree · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n^2)// Space: O(n) // bfs, iterative dfsclass Solution {public:    int maxSum(TreeNode* root) {        const auto& bfs = [&]() {            vector<vector<int>> adj(1);            vector<int> vals = {root->val};            vector<pair<TreeNode*, int>> q = {{root, 0}};            while (!empty(q)) {                vector<pair<TreeNode*, int>> new_q;                for (const auto& [u, p] : q) {                    vals.emplace_back(u->val);                    adj.emplace_back();                    const auto& i = size(adj) - 1;                    if (p != -1) {                        adj[i].emplace_back(p);                        adj[p].emplace_back(i);                    }                    for (const auto& node : {u->left, u->right}) {                        if (!node) {                            continue;                        }                        new_q.emplace_back(node, i);                    }                }                q = move(new_q);            }            return pair(adj, vals);        };         const auto& [adj, vals] = bfs();        const auto iter_dfs = [&](int u) {            int result = numeric_limits<int>::min();            int total = 0;            unordered_set<int> lookup;            vector<tuple<int, int, int>> stk = {{1, u, -1}};            while (!empty(stk)) {                const auto [step, u, p] = stk.back(); stk.pop_back();                if (step == 1) {                    if (lookup.count(vals[u])) {                        continue;                    }                    stk.emplace_back(2, u, p);                    lookup.emplace(vals[u]);                    total += vals[u];                    result = max(result, total);                    for (const auto& v : adj[u]) {                        if (v == p) {                            continue;                        }                        stk.emplace_back(1, v, u);                    }                } else if (step == 2) {                    total -= vals[u];                    lookup.erase(vals[u]);                }            }            return result;         };        int result = numeric_limits<int>::min();        for (int u = 0; u < size(adj); ++u) {            result = max(result, iter_dfs(u));        }        return result;    }}; // Time:  O(n^2)// Space: O(n)// dfsclass Solution2 {public:    int maxSum(TreeNode* root) {        vector<vector<int>> adj;        vector<int> vals;        const auto dfs1 = [&](this auto&& dfs1, TreeNode *u, int p) -> void {            vals.emplace_back(u->val);            adj.emplace_back();            const auto& i = size(adj) - 1;            if (p != -1) {                adj[i].emplace_back(p);                adj[p].emplace_back(i);            }            for (const auto& node : {u->left, u->right}) {                if (!node) {                    continue;                }                dfs1(node, i);            }        };         unordered_set<int> lookup;        const auto dfs2 = [&](this auto&& dfs2, int u, int p) {            if (lookup.count(vals[u])) {                return numeric_limits<int>::min();            }            lookup.emplace(vals[u]);            int mx = 0;            for (const auto& v : adj[u]) {                if (v == p) {                    continue;                }                mx = max(mx, dfs2(v, u));            }            lookup.erase(vals[u]);            return vals[u] + mx;        };         dfs1(root, -1);        int result = numeric_limits<int>::min();        for (int u = 0; u < size(adj); ++u) {            result = max(result, dfs2(u, -1));        }        return result;    }};

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