Evaluation of Expression Tree

Evaluation of Expression Tree

Given a simple expression tree, consisting of basic binary operators i.e., + , – ,* and / and some integers, evaluate the expression tree.

Examples:

Input :
Root node of the below tree


Output :
100

Input :
Root node of the below tree


Output :
110

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As all the operators in the tree are binary hence each node will have either 0 or 2 children. As it can be inferred from the examples above , the integer values would appear at the leaf nodes , while the interior nodes represent the operators.
To evaluate the syntax tree , a recursive approach can be followed .

Algorithm :
Let t be the syntax tree
If  t is not null then
      If t.info is operand then
         Return  t.info
      Else
         A = solve(t.left)
         B = solve(t.right)
         return A operator B
         where operator is the info contained in t

The time complexity would be O(n), as each node is visited once. Below is a C++ program for the same:

 #include <iostream>
 #include <cstdlib>
 using namespace std;

 typedef struct node{
     string s;
     node *left;
     node *right;
     node(string x): s(x), left(NULL), right(NULL){}
 }Node;

 // Utility function to return the integer value
 // of a given string
 int toInt(string s){
     int len = s.length();
     int num = ;
     for(int i = ; i < len; i++){
         num = num *  + (s[i]-'');
     }
     return num;
 }

 // Check which operator to apply
 int calculate(const char *c, int lval, int rval){
     int ans;
     switch(*c){
     case '+': ans = lval + rval; break;
     case '-': ans = lval - rval; break;
     case '*': ans = lval * rval; break;
     case '/': ans = lval / rval; break;
     }
     return ans;
 }

 // This function receives a node of the syntax tree
 // and recursively evaluates it
 int eval(Node *root){
     // empty tree
     if(root == NULL)
         return ;
     // leaf node i.e, an integer
     if(root->left == NULL && root->right == NULL)
         return toInt(root->s);
     // Evaluate left subtree
     int lval = eval(root->left);
     // Evaluate right subtree
     int rval = eval(root->right);
     return calculate((root->s).c_str(), lval, rval);
 }

 int main()
 {
     // create a syntax tree
     node *root = new node("+");
     root->left = new node("*");
     root->left->left = new node("");
     root->left->right = new node("");
     root->right = new node("-");
     root->right->left = new node("");
     root->right->right = new node("");
     cout << eval(root) << endl;

     delete(root);

     root = new node("+");
     root->left = new node("*");
     root->left->left = new node("");
     root->left->right = new node("");
     root->right = new node("-");
     root->right->left = new node("");
     root->right->right = new node("/");
     root->right->right->left = new node("");
     root->right->right->right = new node("");

     cout << eval(root);
     system("pause");
     return ;
 }

100

110

参考：http://www.geeksforgeeks.org/evaluation-of-expression-tree/