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Problem solution · C++

Find Minimum Diameter After Merging Two Trees

Find Minimum Diameter After Merging Two Trees: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
49 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

Direct simulation

For Find Minimum Diameter After Merging Two Trees, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 49 lines of C++ from the credited upstream file 3203.cpp.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind Minimum Diameter After Merging Two Trees · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int minimumDiameterAfterMerge(vector<vector<int>>& edges1,                                vector<vector<int>>& edges2) {    const int diameter1 = getDiameter(edges1);    const int diameter2 = getDiameter(edges2);    const int combinedDiameter = (diameter1 + 1) / 2 + (diameter2 + 1) / 2 + 1;    return max({diameter1, diameter2, combinedDiameter});  }  private:  int getDiameter(const vector<vector<int>>& edges) {    const int n = edges.size() + 1;    vector<vector<int>> graph(n);     for (const vector<int>& edge : edges) {      const int u = edge[0];      const int v = edge[1];      graph[u].push_back(v);      graph[v].push_back(u);    }     int maxDiameter = 0;    maxDepth(graph, 0, /*prev=*/-1, maxDiameter);    return maxDiameter;  }   // Similar to 1522. Diameter of N-Ary Tree  // Returns the maximum depth of the subtree rooted at u.  int maxDepth(const vector<vector<int>>& graph, int u, int prev,               int& maxDiameter) {    int maxSubDepth1 = 0;    int maxSubDepth2 = 0;    for (const int v : graph[u]) {      if (v == prev)        continue;      const int maxSubDepth = maxDepth(graph, v, u, maxDiameter);      if (maxSubDepth > maxSubDepth1) {        maxSubDepth2 = maxSubDepth1;        maxSubDepth1 = maxSubDepth;      } else if (maxSubDepth > maxSubDepth2) {        maxSubDepth2 = maxSubDepth;      }    }    maxDiameter = max(maxDiameter, maxSubDepth1 + maxSubDepth2);    return 1 + maxSubDepth1;  }};

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