【数据结构】中缀表达式构造二叉树转换成后缀表达式

最新推荐文章于 2025-05-24 12:10:20 发布

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最新推荐文章于 2025-05-24 12:10:20 发布

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本文介绍了一个使用 C 语言实现的程序，该程序通过构建表达式树来解析数学表达式3*4+5-2/1。文章详细展示了如何使用栈来辅助构造表达式树，并提供了完整的源代码及展示如何遍历这棵树以输出表达式的具体过程。

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#include "stdio.h"  
#include "string.h"  
#include "stdlib.h"  
#include "stack"  
using namespace std;  
  
const char str[] = "3*4+5-2/1";  
  
struct tree  
{  
    char c;  
    struct tree* left;  
    struct tree* right;  
};  
stack<struct tree*> subTreeStack;  
stack<char> operatorStack;  
  
  
struct tree*  BuildTree()  
{   
  
    struct tree* node;  
    for (unsigned int pos = 0; pos < strlen(str); pos++)  
    {  
        if (str[pos] - '0' >= 0 && str[pos] - '0' <= 9)    //若是数字则作为叶子节点  
        {  
  
            node = (struct tree*) malloc(sizeof(struct tree));  
            node->c = str[pos];  
            node->left = NULL;  
            node->right = NULL;  
              
          
              
            subTreeStack.push(node);  
        }  
        else if (str[pos] == '*' || str[pos] == '/' || str[pos] == '+' || str[pos] == '-')  
        {  
            if (!operatorStack.empty() && (str[pos] == '+' || str[pos] == '-'))  //若优先级比栈顶元素低  
            {  
                node = (struct tree*) malloc(sizeof(struct tree));  
                node->c = operatorStack.top();  
                node->right = subTreeStack.top();  
                subTreeStack.pop();  
                node->left = subTreeStack.top();  
                subTreeStack.pop();  
                  
                subTreeStack.push(node);  
              
  
                operatorStack.pop();  
                operatorStack.push(str[pos]);  
            }  
            else  
                operatorStack.push(str[pos]);  
        }  
      
      
    }  
      
      
    while(!operatorStack.empty())  
    {  
      
        node = (struct tree*) malloc(sizeof(struct tree));  
        node->c = operatorStack.top();  
        operatorStack.pop();  
        node->right = subTreeStack.top();  
        subTreeStack.pop();  
        node->left = subTreeStack.top();  
        subTreeStack.pop();  
          
        subTreeStack.push(node);  
      
    }  
    return subTreeStack.top();  
  
}  
void PrintTree(struct tree* node)  
{  
    if (node == NULL)  
        return;  
    PrintTree(node->left);  
    printf("%c", node->c);  
    PrintTree(node->right);  
}  
  
  
void main()  
{  
    struct tree* root = BuildTree();  
    PrintTree(root);  
}