用C++写一个输入任意中缀表达式转换抽象语法树代码

以下是使用C++实现输入任意中缀表达式转换抽象语法树的示例代码：

#include <iostream>
#include <stack>
#include <string>
#include <sstream>

using namespace std;

struct TreeNode {
    string val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(string v): val(v), left(NULL), right(NULL) {}
};

bool isOperator(string s) {
    return s == "+" || s == "-" || s == "*" || s == "/";
}

int getPriority(string op) {
    if (op == "+" || op == "-") {
        return 1;
    } else if (op == "*" || op == "/") {
        return 2;
    } else {
        return 0;
    }
}

TreeNode* buildAST(string expression) {
    stack<TreeNode*> nodeStack;
    stack<string> opStack;
    stringstream ss(expression);
    string token;
    while (ss >> token) {
        if (isOperator(token)) {
            while (!opStack.empty() && getPriority(opStack.top()) >= getPriority(token)) {
                TreeNode* right = nodeStack.top();
                nodeStack.pop();
                TreeNode* left = nodeStack.top();
                nodeStack.pop();
                string op = opStack.top();
                opStack.pop();
                TreeNode* newNode = new TreeNode(op);
                newNode->left = left;
                newNode->right = right;
                nodeStack.push(newNode);
            }
            opStack.push(token);
        } else {
            TreeNode* newNode = new TreeNode(token);
            nodeStack.push(newNode);
        }
    }
    while (!opStack.empty()) {
        TreeNode* right = nodeStack.top();
        nodeStack.pop();
        TreeNode* left = nodeStack.top();
        nodeStack.pop();
        string op = opStack.top();
        opStack.pop();
        TreeNode* newNode = new TreeNode(op);
        newNode->left = left;
        newNode->right = right;
        nodeStack.push(newNode);
    }
    return nodeStack.top();
}

void inorder(TreeNode* root) {
    if (root == NULL) return;
    if (root->left != NULL) {
        cout << "(";
        inorder(root->left);
    }
    cout << root->val;
    if (root->right != NULL) {
        inorder(root->right);
        cout << ")";
    }
}

int main() {
    string expression = "3 + 4 * 5 - 6 / 2";
    TreeNode* root = buildAST(expression);
    inorder(root);
    cout << endl;
    return 0;
}

该示例代码实现了输入任意中缀表达式，将其转换为抽象语法树，并输出其对应的中缀表达式形式。其中，`buildAST`函数实现了将中缀表达式转换为抽象语法树的过程，`inorder`函数实现了中序遍历抽象语法树并输出中缀表达式的过程。