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Evaluarea arborelui de expresie

Dat un simplu arbore de expresie constând din operatori binari de bază, adică + - * și / și unele numere întregi evaluează arborele de expresie.

Exemple:



Intrare: Nodul rădăcină al arborelui de mai jos

poza 2' title=

Ieșire: 100



Intrare: Nodul rădăcină al arborelui de mai jos

un exemplu de sistem open source este

poza1' title=

Ieșire: 110



Practică recomandată Arborele de expresie Încearcă!

Abordare: Abordarea pentru rezolvarea acestei probleme se bazează pe următoarele observații:

Deoarece toți operatorii din arbore sunt binari, fiecare nod va avea fie 0, fie 2 copii. După cum se poate deduce din exemplele de mai sus, valorile întregi ar apărea la nodurile frunză, în timp ce nodurile interioare reprezintă operatorii.

Prin urmare, putem face parcurgerea în ordine a arborelui binar și evaluăm expresia pe măsură ce avansăm.

Pentru a evalua arborele de sintaxă poate fi urmată o abordare recursivă.

Algoritm:

  • Fie t arborele de sintaxă
  • Dacă  t nu este nul, atunci      
    • Dacă t.info este operand atunci  
      • Reveniți  t.info
    • Altfel
      • A = rezolva (t.stânga)
      • B = rezolva (t.dreapta)
      • return A operator B unde operator este informația conținută în t

Mai jos este implementarea abordării de mai sus:

C++
// C++ program to evaluate an expression tree  #include     using namespace std;  // Class to represent the nodes of syntax tree  class node  {  public:   string info;   node *left = NULL *right = NULL;   node(string x)   {   info = x;   }  };  // Utility function to return the integer value  // of a given string  int toInt(string s)  {   int num = 0;     // Check if the integral value is   // negative or not   // If it is not negative generate the number   // normally   if(s[0]!='-')   for (int i=0; i<s.length(); i++)   num = num*10 + (int(s[i])-48);   // If it is negative calculate the +ve number   // first ignoring the sign and invert the   // sign at the end   else  {   for (int i=1; i<s.length(); i++)   num = num*10 + (int(s[i])-48);   num = num*-1;   }     return num;  }  // This function receives a node of the syntax tree  // and recursively evaluates it  int eval(node* root)  {   // empty tree   if (!root)   return 0;   // leaf node i.e an integer   if (!root->left && !root->right)   return toInt(root->info);   // Evaluate left subtree   int l_val = eval(root->left);   // Evaluate right subtree   int r_val = eval(root->right);   // Check which operator to apply   if (root->info=='+')   return l_val+r_val;   if (root->info=='-')   return l_val-r_val;   if (root->info=='*')   return l_val*r_val;   return l_val/r_val;  }  //driver function to check the above program  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);   return 0;  }  
Java
// Java program to evaluate expression tree import java.lang.*; class GFG{   Node root; // Class to represent the nodes of syntax tree public static class Node  {  String data;  Node left right;  Node(String d)  {  data = d;  left = null;  right = null;  } } private static int toInt(String s) {  int num = 0;  // Check if the integral value is  // negative or not  // If it is not negative generate   // the number normally  if (s.charAt(0) != '-')  for(int i = 0; i < s.length(); i++)  num = num * 10 + ((int)s.charAt(i) - 48);    // If it is negative calculate the +ve number  // first ignoring the sign and invert the  // sign at the end  else  {  for(int i = 1; i < s.length(); i++)   num = num * 10 + ((int)(s.charAt(i)) - 48);  num = num * -1;  }  return num; } // This function receives a node of the syntax // tree and recursively evaluate it public static int evalTree(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.data);  // Evaluate left subtree  int leftEval = evalTree(root.left);  // Evaluate right subtree  int rightEval = evalTree(root.right);  // Check which operator to apply  if (root.data.equals('+'))  return leftEval + rightEval;  if (root.data.equals('-'))  return leftEval - rightEval;  if (root.data.equals('*'))  return leftEval * rightEval;  return leftEval / rightEval; } // Driver code public static void main(String[] args) {    // Creating a sample 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');  System.out.println(evalTree(root));  root = null;  // Creating a sample tree  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');  System.out.println(evalTree(root)); } } // This code is contributed by Ankit Gupta 
Python3
# Python program to evaluate expression tree # Class to represent the nodes of syntax tree class node: def __init__(self value): self.left = None self.data = value self.right = None # This function receives a node of the syntax tree # and recursively evaluate it def evaluateExpressionTree(root): # empty tree if root is None: return 0 # leaf node if root.left is None and root.right is None: return int(root.data) # evaluate left tree left_sum = evaluateExpressionTree(root.left) # evaluate right tree right_sum = evaluateExpressionTree(root.right) # check which operation to apply if root.data == '+': return left_sum + right_sum elif root.data == '-': return left_sum - right_sum elif root.data == '*': return left_sum * right_sum else: return left_sum // right_sum # Driver function to test above problem if __name__ == '__main__': # creating a sample tree root = node('+') root.left = node('*') root.left.left = node('5') root.left.right = node('-4') root.right = node('-') root.right.left = node('100') root.right.right = node('20') print (evaluateExpressionTree(root)) root = None # creating a sample tree root = node('+') root.left = node('*') root.left.left = node('5') root.left.right = node('4') root.right = node('-') root.right.left = node('100') root.right.right = node('/') root.right.right.left = node('20') root.right.right.right = node('2') print (evaluateExpressionTree(root)) # This code is contributed by Harshit Sidhwa 
C#
// C# program to evaluate expression tree using System; public class GFG  {  // Class to represent the nodes of syntax tree  public class Node {  public  String data;  public  Node left right;  public Node(String d) {  data = d;  left = null;  right = null;  }  }  private static int toInt(String s) {  int num = 0;  // Check if the integral value is  // negative or not  // If it is not negative generate  // the number normally  if (s[0] != '-')  for (int i = 0; i < s.Length; i++)  num = num * 10 + ((int) s[i] - 48);  // If it is negative calculate the +ve number  // first ignoring the sign and invert the  // sign at the end  else {  for (int i = 1; i < s.Length; i++)  num = num * 10 + ((int) (s[i]) - 48);  num = num * -1;  }  return num;  }  // This function receives a node of the syntax  // tree and recursively evaluate it  public static int evalTree(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.data);  // Evaluate left subtree  int leftEval = evalTree(root.left);  // Evaluate right subtree  int rightEval = evalTree(root.right);  // Check which operator to apply  if (root.data.Equals('+'))  return leftEval + rightEval;  if (root.data.Equals('-'))  return leftEval - rightEval;  if (root.data.Equals('*'))  return leftEval * rightEval;  return leftEval / rightEval;  }  // Driver code  public static void Main(String[] args) {  // Creating a sample 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');  Console.WriteLine(evalTree(root));  root = null;  // Creating a sample tree  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');  Console.WriteLine(evalTree(root));  } } // This code is contributed by umadevi9616  
JavaScript
<script> // javascript program to evaluate expression tree  var root;  // Class to represent the nodes of syntax tree  class Node {  constructor(val) {  this.data = val;  this.left = null;  this.right = null;  }  }  function toInt( s) {  var num = 0;    // Check if the integral value is  // negative or not  // If it is not negative generate  // the number normally  if (s.charAt(0) != '-')  for (i = 0; i < s.length; i++)  num = num * 10 + ( s.charCodeAt(i) - 48);  // If it is negative calculate the +ve number  // first ignoring the sign and invert the  // sign at the end  else {  for (i = 1; i < s.length; i++)  num = num * 10 + (s.charCodeAt(i) - 48);  num = num * -1;  }  return num;  }  // This function receives a node of the syntax  // tree and recursively evaluate it  function evalTree(root) {  // Empty tree  if (root == null)  return 0;  // Leaf node i.e an integer  if (root.left == null && root.right == null)  return toInt(root.data);  // Evaluate left subtree  var leftEval = evalTree(root.left);  // Evaluate right subtree  var rightEval = evalTree(root.right);  // Check which operator to apply  if (root.data === ('+'))  return leftEval + rightEval;  if (root.data === ('-'))  return leftEval - rightEval;  if (root.data === ('*'))  return leftEval * rightEval;  return leftEval / rightEval;  }  // Driver code    // Creating a sample tree  var 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');  document.write(evalTree(root));  root = null;  // Creating a sample tree  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');  document.write('  
'
+evalTree(root)); // This code is contributed by gauravrajput1 </script>

Ieșire
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Complexitatea timpului: O(n) deoarece fiecare nod este vizitat o dată.
Spațiu auxiliar: Pe)