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Building Expression tree from Prefix Expression

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  • Difficulty Level : Hard
  • Last Updated : 03 Nov, 2021

Given a character array a[] representing a prefix expression. The task is to build an Expression Tree for the expression and then print the infix and postfix expression of the built tree.
Examples: 
 

Input: a[] = “*+ab-cd” 
Output: The Infix expression is: 
a + b * c – d 
The Postfix expression is: 
a b + c d – *
Input: a[] = “+ab” 
Output: The Infix expression is: 
a + b 
The Postfix expression is: 
a b + 
 

 

Approach: If the character is an operand i.e. X then it’ll be the leaf node of the required tree as all the operands are at the leaf in an expression tree. Else if the character is an operator and of the form OP X Y then it’ll be an internal node with left child as the expressionTree(X) and right child as the expressionTree(Y) which can be solved using a recursive function.
Below is the implementation of the above approach:
 

C




// C program to construct an expression tree
// from prefix expression
#include <stdio.h>
#include <stdlib.h>
 
// Represents a node of the required tree
typedef struct node {
    char data;
    struct node *left, *right;
} node;
 
// Function to recursively build the expression tree
char* add(node** p, char* a)
{
 
    // If its the end of the expression
    if (*a == '\0')
        return '\0';
 
    while (1) {
        char* q = "null";
        if (*p == NULL) {
 
            // Create a node with *a as the data and
            // both the children set to null
            node* nn = (node*)malloc(sizeof(node));
            nn->data = *a;
            nn->left = NULL;
            nn->right = NULL;
            *p = nn;
        }
        else {
 
            // If the character is an operand
            if (*a >= 'a' && *a <= 'z') {
                return a;
            }
 
            // Build the left sub-tree
            q = add(&(*p)->left, a + 1);
 
            // Build the right sub-tree
            q = add(&(*p)->right, q + 1);
 
            return q;
        }
    }
}
 
// Function to print the infix expression for the tree
void inr(node* p) // recursion
{
    if (p == NULL) {
        return;
    }
    else {
        inr(p->left);
        printf("%c ", p->data);
        inr(p->right);
    }
}
 
// Function to print the postfix expression for the tree
void postr(node* p)
{
    if (p == NULL) {
        return;
    }
    else {
        postr(p->left);
        postr(p->right);
        printf("%c ", p->data);
    }
}
 
// Driver code
int main()
{
    node* s = NULL;
    char a[] = "*+ab-cd";
    add(&s, a);
    printf("The Infix expression is:\n ");
    inr(s);
    printf("\n");
    printf("The Postfix expression is:\n ");
    postr(s);
    return 0;
}

Python3




# Python3 program to construct an expression tree
# from prefix expression
 
# Represents a node of the required tree
class node:
    def __init__(self,c):
        self.data=c
        self.left=self.right=None
 
# Function to recursively build the expression tree
def add(a):
 
    # If its the end of the expression
    if (a == ''):
        return ''
 
    # If the character is an operand
    if a[0]>='a' and a[0]<='z':
        return node(a[0]),a[1:]
    else:
        # Create a node with a[0] as the data and
        # both the children set to null
        p=node(a[0])
        # Build the left sub-tree
        p.left,q=add(a[1:])
        # Build the right sub-tree
        p.right,q=add(q)
        return p,q
         
 
# Function to print the infix expression for the tree
def inr(p): #recursion
 
    if (p == None):
        return
    else:
        inr(p.left)
        print(p.data,end=' ')
        inr(p.right)
 
# Function to print the postfix expression for the tree
def postr(p):
 
    if (p == None):
        return
    else:
        postr(p.left)
        postr(p.right)
        print(p.data,end=' ')
 
# Driver code
if __name__ == '__main__':
     
    a = "*+ab-cd"
    s,a=add(a)
    print("The Infix expression is:")
    inr(s)
    print()
    print("The Postfix expression is:")
    postr(s)
 
# This code is contributed by Amartya Ghosh

Output: 

The Infix expression is:
 a + b * c - d 
The Postfix expression is:
 a b + c d - *

 


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