Expression Tree
The expression tree is a binary tree in which each internal node corresponds to the operator and each leaf node corresponds to the operand so for example expression tree for 3 + ((5+9)*2) would be:
Inorder traversal of expression tree produces infix version of given postfix expression (same with postorder traversal it gives postfix expression)
Evaluating the expression represented by an expression tree:
Let t be the expression tree
If t is not null then
If t.value is operand then
Return t.value
A = solve(t.left)
B = solve(t.right)
// calculate applies operator 't.value'
// on A and B, and returns value
Return calculate(A, B, t.value)Construction of Expression Tree:
Now For constructing an expression tree we use a stack. We loop through input expression and do the following for every character.
- If a character is an operand push that into the stack
- If a character is an operator pop two values from the stack make them its child and push the current node again.
In the end, the only element of the stack will be the root of an expression tree.
Below is the implementation of the above approach:
C++
// C++ program for expression tree#include<bits/stdc++.h>using namespace std;// An expression tree nodestruct et{ char value; et* left, *right;};// A utility function to check if 'c'// is an operatorbool isOperator(char c){ if (c == '+' || c == '-' || c == '*' || c == '/' || c == '^') return true; return false;}// Utility function to do inorder traversalvoid inorder(et *t){ if(t) { inorder(t->left); printf("%c ", t->value); inorder(t->right); }}// A utility function to create a new nodeet* newNode(char v){ et *temp = new et; temp->left = temp->right = NULL; temp->value = v; return temp;};// Returns root of constructed tree for given// postfix expressionet* constructTree(char postfix[]){ stack<et *> st; et *t, *t1, *t2; // Traverse through every character of // input expression for (int i=0; i<strlen(postfix); i++) { // If operand, simply push into stack if (!isOperator(postfix[i])) { t = newNode(postfix[i]); st.push(t); } else // operator { t = newNode(postfix[i]); // Pop two top nodes t1 = st.top(); // Store top st.pop(); // Remove top t2 = st.top(); st.pop(); // make them children t->right = t1; t->left = t2; // Add this subexpression to stack st.push(t); } } // only element will be root of expression // tree t = st.top(); st.pop(); return t;}// Driver program to test aboveint main(){ char postfix[] = "ab+ef*g*-"; et* r = constructTree(postfix); printf("infix expression is \n"); inorder(r); return 0;} |
Java
// Java program to construct an expression treeimport java.util.Stack;// Java program for expression treeclass Node { char value; Node left, right; Node(char item) { value = item; left = right = null; }}class ExpressionTree { // A utility function to check if 'c' // is an operator boolean isOperator(char c) { if (c == '+' || c == '-' || c == '*' || c == '/' || c == '^') { return true; } return false; } // Utility function to do inorder traversal void inorder(Node t) { if (t != null) { inorder(t.left); System.out.print(t.value + " "); inorder(t.right); } } // Returns root of constructed tree for given // postfix expression Node constructTree(char postfix[]) { Stack<Node> st = new Stack<Node>(); Node t, t1, t2; // Traverse through every character of // input expression for (int i = 0; i < postfix.length; i++) { // If operand, simply push into stack if (!isOperator(postfix[i])) { t = new Node(postfix[i]); st.push(t); } else // operator { t = new Node(postfix[i]); // Pop two top nodes // Store top t1 = st.pop(); // Remove top t2 = st.pop(); // make them children t.right = t1; t.left = t2; // System.out.println(t1 + "" + t2); // Add this subexpression to stack st.push(t); } } // only element will be root of expression // tree t = st.peek(); st.pop(); return t; } public static void main(String args[]) { ExpressionTree et = new ExpressionTree(); String postfix = "ab+ef*g*-"; char[] charArray = postfix.toCharArray(); Node root = et.constructTree(charArray); System.out.println("infix expression is"); et.inorder(root); }}// This code has been contributed by Mayank Jaiswal |
Python
# Python program for expression tree# An expression tree nodeclass Et: # Constructor to create a node def __init__(self , value): self.value = value self.left = None self.right = None# A utility function to check if 'c'# is an operatordef isOperator(c): if (c == '+' or c == '-' or c == '*' or c == '/' or c == '^'): return True else: return False# A utility function to do inorder traversaldef inorder(t): if t is not None: inorder(t.left) print t.value, inorder(t.right)# Returns root of constructed tree for# given postfix expressiondef constructTree(postfix): stack = [] # Traverse through every character of input expression for char in postfix : # if operand, simply push into stack if not isOperator(char): t = Et(char) stack.append(t) # Operator else: # Pop two top nodes t = Et(char) t1 = stack.pop() t2 = stack.pop() # make them children t.right = t1 t.left = t2 # Add this subexpression to stack stack.append(t) # Only element will be the root of expression tree t = stack.pop() return t# Driver program to test abovepostfix = "ab+ef*g*-"r = constructTree(postfix)print "Infix expression is"inorder(r) |
C#
// C# program to construct an expression treeusing System;using System.Collections;public class Node{ public char value; public Node left, right; public Node(char item) { value = item; left = right = null; }}class ExpressionTree{ // A utility function to check if 'c' // is an operator Boolean isOperator(char c) { if (c == '+' || c == '-' || c == '*' || c == '/' || c == '^') { return true; } return false; } // Utility function to do inorder traversal void inorder(Node t) { if (t != null) { inorder(t.left); Console.Write(t.value + " "); inorder(t.right); } } // Returns root of constructed tree for given // postfix expression Node constructTree(char []postfix) { Stack st = new Stack(); Node t, t1, t2; // Traverse through every character of // input expression for (int i = 0; i < postfix.Length; i++) { // If operand, simply Push into stack if (!isOperator(postfix[i])) { t = new Node(postfix[i]); st.Push(t); } else // operator { t = new Node(postfix[i]); // Pop two top nodes // Store top t1 = (Node)st.Pop(); // Remove top t2 = (Node)st.Pop(); // make them children t.right = t1; t.left = t2; // System.out.println(t1 + "" + t2); // Add this subexpression to stack st.Push(t); } } // only element will be root of // expression tree t = (Node)st.Peek(); st.Pop(); return t; } // Driver code public static void Main(String []args) { ExpressionTree et = new ExpressionTree(); String postfix = "ab+ef*g*-"; char[] charArray = postfix.ToCharArray(); Node root = et.constructTree(charArray); Console.WriteLine("infix expression is"); et.inorder(root); }}// This code has been contributed by Arnab Kundu |
Output
infix expression is a + b - e * f * g
This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
