# Evaluation of Expression Tree

Source: Internet
Author: User

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 Treeoutput:100input:root node of the below treeoutput:110`

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As all of the operators in the tree is binary hence each node would 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 nod ES represent the operators.
To evaluate the syntax tree, a recursive approach can be followed.

`Algorithm:let t be the syntax Treeif  t was not null then      If t.info are operand then           Return  t.info      else< C6/>a = Solve (t.left)         B = Solve (t.right)         return A operator B         where operator is the info contained in T`

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

`1#include <iostream>2#include <cstdlib>3 using namespacestd;4 5typedefstructnode{6     strings;7Node *Left ;8Node *Right ;9Nodestringx): S (x), left (null), right (null) {}Ten }node; One  A //Utility function to return the integer value - //of a given string - intToInt (strings) { the     intLen =s.length (); -     intnum =0; -      for(inti =0; i < Len; i++){ -num = num *Ten+ (s[i]-'0'); +     } -     returnnum; + } A  at //Check which operator to apply - intCalculateConst Char*c,intLval,intrval) { -     intans; -     Switch(*c) { -      Case '+': ans = lval + rval; Break; -      Case '-': ans = lval-rval; Break; in      Case '*': ans = lval * rval; Break; -      Case '/': ans = lval/rval; Break; to     } +     returnans; - } the  * //This function receives a node of the syntax tree \$ //and recursively evaluates itPanax Notoginseng intEval (Node *root) { -     //Empty Tree the     if(Root = =NULL) +         return 0; A     //leaf node i.e, an integer the     if(Root->left = = NULL && Root->right = =NULL) +         returnToInt (root->s); -     //Evaluate left subtree \$     intLval = eval (root->Left ); \$     //Evaluate Right subtree -     intRval = eval (root->Right ); -     returnCalculate ((root->s). C_str (), lval, rval); the } - Wuyi intMain () the { -     //Create a syntax tree WuNode *root =NewNode"+"); -Root->left =NewNode"*"); AboutRoot->left->left =NewNode"5"); \$Root->left->right =NewNode"4"); -Root->right =NewNode"-"); -Root->right->left =NewNode" -"); -Root->right->right =NewNode" -"); Acout << eval (root) <<Endl; +   the     Delete(root); -   \$Root =NewNode"+"); theRoot->left =NewNode"*"); theRoot->left->left =NewNode"5"); theRoot->left->right =NewNode"4"); theRoot->right =NewNode"-"); -Root->right->left =NewNode" -"); inRoot->right->right =NewNode"/"); theRoot->right->right->left =NewNode" -"); theRoot->right->right->right =NewNode"2"); About   thecout <<eval (root); theSystem"Pause"); the     return 0; +}`

100

110

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

Evaluation of Expression Tree

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