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Problem solution · C++

Design an Expression Tree With Evaluate Function

Design an Expression Tree With Evaluate Function: a C++ solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
62 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

Sliding window or two pointers

For Design an Expression Tree With Evaluate Function, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 62 lines of C++ from the credited upstream file 1628.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 1 loop block detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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 codeDesign an Expression Tree With Evaluate Function · C++C++
Use this to learn the idea, then write your own version.
/** * This is the interface for the expression tree Node. * You should not remove it, and you can define some classes to implement it. */ class Node { public:  virtual ~Node() {};  virtual int evaluate() const = 0;  protected:  // define your fields here}; class ExpNode : public Node { public:  ExpNode(const string& val, ExpNode* left, ExpNode* right)      : val(val), left(left), right(right) {}   int evaluate() const override {    return left == nullptr && right == nullptr               ? stoi(val)               : op.at(val)(left->evaluate(), right->evaluate());  }  private:  static const inline unordered_map<string, function<long(long, long)>> op{      {"+", std::plus<long>()},      {"-", std::minus<long>()},      {"*", std::multiplies<long>()},      {"/", std::divides<long>()}};  const string val;  const ExpNode* const left;  const ExpNode* const right;}; /** * This is the TreeBuilder class. * You can treat it as the driver code that takes the postinfix input * and returns the expression tree represnting it as a Node. */ class TreeBuilder { public:  Node* buildTree(vector<string>& postfix) {    stack<ExpNode*> stack;     for (const string& val : postfix)      if (val == "+" || val == "-" || val == "*" || val == "/") {        ExpNode* right = stack.top();        stack.pop();        ExpNode* left = stack.top();        stack.pop();        stack.push(new ExpNode(val, left, right));      } else {        stack.push(new ExpNode(val, nullptr, nullptr));      }     return stack.top();  }};

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