Build a binary tree from an infix expression without using a stack

Question

Recently I wrote an algorithm to convert an infix expression to a binary tree without using any stack . However, as I search on web, I find the algorithms described there are all based on stack(or recursion).

So I begin to worry about the correctness about my algorithm, though I cannot prove it's incorrect yet.

Question

Do you know whether it's technically possible to convert it without any stack or not? Is my algorithm wrong?

Short description

It's based on:

1. An operand in an infix expression belongs to either the right child of the operator in front of it, or the left child of the operator behind it.

2. If an operator OP2 has higher precedence than its preceding operator OP1 , the previous operand x becomes the left child of OP2 , and OP2 becomes the right child of OP1 .

3. If an operator OP2 has lower precedence than its preceding operator OP1 , the previous operand x becomes the right child of OP1 . Go up the tree from OP1 , compare the precedence of each ancestor of OP1 with that of OP2 until OP2 <= ancestor OP . Then OP2 becomes the right child of OP .

The program

#include <iostream>
#include <string>
#include <sstream>
#include <cassert>

using namespace std;

typedef struct Node{
   // store operator or operand
   string data;
   // only valid for operator
   int precedence;
   struct Node* parent;
   struct Node* left;
   struct Node* right;
}CNode, *PNode;

PNode CreateNode(const string& x)
{
   PNode p = new CNode;
   p->parent = p->left = p->right = NULL;
   p->data = x;
   return p;
}

bool IsOperator(const string& x)
{
   // Since the only impact of parentheses () is on precedence, 
   // they are not considered as operators here
   return ((x.length() == 1) &&
           (x[0] == '*' ||
            x[0] == '/' ||
            x[0] == '+' ||
            x[0] == '-'));
}

bool IsLeftParenthesis(const string& x)
{
   return x == "(";
}

bool IsRightParenthesis(const string& x)
{
   return x == ")";
}

bool IsOperand(const string& x)
{
   int y;
   stringstream ss(x);
   if (ss >> y) return true;
   else return false;
}

int GetPrecedence(const string& x)
{
   assert(IsOperator(x));
   if (x[0] == '*' || x[0] == '/') return 2;
   else return 1;
}

PNode CreateInfixTree(const string& exp)
{
   // create a dummy root with minimal precedence
   // its content is trivial
   PNode root = CreateNode("0");
   root->precedence = INT_MIN;

   // the previous operand of current operator
   PNode preOperand = NULL;
   // the previous operator of current operator
   PNode preOperator = root;
   // the impact of preceding parenthesis, if any
   int correction = 0;

   string token;
   stringstream ss(exp);

   while (ss >> token)
   {
      if (IsOperand(token))
      {
         preOperand = CreateNode(token);
      }
      else if (IsOperator(token))
      {
         PNode p = CreateNode(token);
         p->precedence = GetPrecedence(token) + correction;
         if (p->precedence > preOperator->precedence)
         {
            p->left = preOperand;
            preOperator->right = p;
            p->parent = preOperator;
         }
         else
         {
            preOperator->right = preOperand;
            PNode q = preOperator->parent;
            while (p->precedence <= q->precedence) q = q->parent;

            p->left = q->right;
            q->right = p;
            p->parent = q;
         }
         preOperand = NULL;
         preOperator = p;

      }//else if (IsOperator(token)
      else if (IsLeftParenthesis(token))
      {
         correction += 2;
      }
      else if (IsRightParenthesis(token))
      {
         correction -= 2;
      }
      else
      {
         cout << "illegal token found: " << token << endl;
         break;
      }
   }//while

   if (preOperand == NULL)
       cout << "illegal expression: cannot end with operator: "
            << preOperator->data << endl;
   else preOperator->right = preOperand;

   // delete dummy root
   PNode realRoot = root->right;
   delete root;
   if (realRoot) realRoot->parent = NULL;
   return realRoot;
}

void PostOrderPrintTree(PNode node)
{
   if (node)
   {
      PostOrderPrintTree(node->left);
      PostOrderPrintTree(node->right);
      cout << node->data << " ";
   }
}

int main()
{
   // valid operators: + - * / ( )
   // valid operands: integers
   // whitespace separated as: ( 1 + 2 ) * 3
   string exp;
   getline(cin, exp);
   PNode root = CreateInfixTree(exp);
   PostOrderPrintTree(root);
   cout << endl;
}

Answer 1

这是你的堆栈：

while (p->precedence <= q->precedence) q = q->parent;