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 = 0;
    for(int i = 0; i < len; i++){
        num = num * 10 + (s[i]-'0');
    }
    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 0;
    // 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("5");
    root->left->right = new node("4");
    root->right = new node("-");
    root->right->left = new node("100");
    root->right->right = new node("20");
    cout << eval(root) << endl;
 
    delete(root);
 
    root = new node("+");
    root->left = new node("*");
    root->left->left = new node("5");
    root->left->right = new node("4");
    root->right = new node("-");
    root->right->left = new node("100");
    root->right->right = new node("/");
    root->right->right->left = new node("20");
    root->right->right->right = new node("2");
 
    cout << eval(root);
    system("pause");
    return 0;
}

100

110

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