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Problem solution · C++

Minimum Flips in Binary Tree to Get Result

Minimum Flips in Binary Tree to Get Result: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
44 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

Dynamic programming

For Minimum Flips in Binary Tree to Get Result, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 44 lines of C++ from the credited upstream file 2313.cpp.
  • The implementation visibly relies on sequence storage, hash lookup, cached states.
  • 1 loop block detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 codeMinimum Flips in Binary Tree to Get Result · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int minimumFlips(TreeNode* root, bool result) {    return dp(root, result);  }  private:  struct PairHash {    template <class T1, class T2>    std::size_t operator()(const std::pair<T1, T2>& p) const {      return std::hash<T1>{}(p.first) ^ std::hash<T2>{}(p.second);    }  };   unordered_map<pair<TreeNode*, bool>, int, PairHash> mem;   // Returns the minimum flips to make the subtree become the target.  int dp(TreeNode* root, bool target) {    const pair<TreeNode*, bool> key{root, target};    if (const auto it = mem.find(key); it != mem.cend())      return it->second;    if (root->val == 0 || root->val == 1)  // the leaf      return root->val == target ? 0 : 1;    if (root->val == 5)  // NOT      return dp(root->left == nullptr ? root->right : root->left, !target);     vector<pair<int, int>> nextTargets;    if (root->val == 2)  // OR      nextTargets = target ? vector<pair<int, int>>{{0, 1}, {1, 0}, {1, 1}}                           : vector<pair<int, int>>{{0, 0}};    else if (root->val == 3)  // AND      nextTargets = target ? vector<pair<int, int>>{{1, 1}}                           : vector<pair<int, int>>{{0, 0}, {0, 1}, {1, 0}};    else  // root.val == 4 (XOR)      nextTargets = target ? vector<pair<int, int>>{{0, 1}, {1, 0}}                           : vector<pair<int, int>>{{0, 0}, {1, 1}};     int ans = INT_MAX;    for (const auto& [leftTarget, rightTarget] : nextTargets)      ans = min(ans, dp(root->left, leftTarget) + dp(root->right, rightTarget));    return mem[key] = ans;  }};

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