Building Expression tree from Prefix Expression

Last Updated : 22 Dec, 2023

Given a character array a[] represents 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++ code for the above approach
#include <iostream>
#include <cstring>
#include <cstdlib>
 
using namespace std;
 
// Represents a node of the required tree
struct node {
    char data;
    node *left, *right;
};
 
// 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 == nullptr) {
 
            // Create a node with *a as the data and
            // both the children set to null
            node* nn = new node;
            nn->data = *a;
            nn->left = nullptr;
            nn->right = nullptr;
            *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 == nullptr) {
        return;
    }
    else {
        inr(p->left);
        cout << p->data << " ";
        inr(p->right);
    }
}
 
// Function to print the postfix expression for the tree
void postr(node* p)
{
    if (p == nullptr) {
        return;
    }
    else {
        postr(p->left);
        postr(p->right);
        cout << p->data << " ";
    }
}
 
int main()
{
    node* s = nullptr;
    char a[] = "*+ab-cd";
    add(&s, a);
    cout << "The Infix expression is:\n ";
    inr(s);
    cout << "\n";
    cout << "The Postfix expression is:\n ";
    postr(s);
    return 0;
}
//This code is contributed by Potta Lokesh

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;
}

Java

import java.util.Stack;
 
// Represents a node of the required tree
class Node {
    char data;
    Node left, right;
 
    public Node(char item) {
        data = item;
        left = right = null;
    }
}
 
class ExpressionTree {
 
    static boolean isOperator(char c) {
        return c == '+' || c == '-' || c == '*' || c == '/';
    }
 
    static Node buildTree(String prefix) {
        Stack<Node> stack = new Stack<Node>();
 
        // Traverse the prefix expression in reverse order
        for (int i = prefix.length() - 1; i >= 0; i--) {
            char c = prefix.charAt(i);
 
            if (isOperator(c)) {
                Node left = stack.pop();
                Node right = stack.pop();
                Node node = new Node(c);
                node.left = left;
                node.right = right;
                stack.push(node);
            } else {
                Node node = new Node(c);
                stack.push(node);
            }
        }
 
        Node root = stack.peek();
        stack.pop();
 
        return root;
    }
 
    // Function to print the infix expression for the tree
    static void inOrder(Node node) {
        if (node != null) {
            inOrder(node.left);
            System.out.print(node.data + " ");
            inOrder(node.right);
        }
    }
 
    // Function to print the postfix expression for the tree
    static void postOrder(Node node) {
        if (node != null) {
            postOrder(node.left);
            postOrder(node.right);
            System.out.print(node.data + " ");
        }
    }
 
    public static void main(String[] args) {
        String prefix = "*+ab-cd";
        Node root = buildTree(prefix);
 
        System.out.println("Infix expression is:");
        inOrder(root);
 
        System.out.println("\nPostfix expression is:");
        postOrder(root);
    }
}

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

C#

using System;
using System.Collections.Generic;
 
// Represents a node of the required tree
class Node
{
    public char data;
    public Node left, right;
 
    public Node(char item)
    {
        data = item;
        left = right = null;
    }
}
 
class ExpressionTree
{
    static bool IsOperator(char c)
    {
        return c == '+' || c == '-' || c == '*' || c == '/';
    }
 
    static Node BuildTree(string prefix)
    {
        Stack<Node> stack = new Stack<Node>();
 
        // Traverse the prefix expression in reverse order
        for (int i = prefix.Length - 1; i >= 0; i--)
        {
            char c = prefix[i];
 
            if (IsOperator(c))
            {
                Node left = stack.Pop();
                Node right = stack.Pop();
                Node node = new Node(c);
                node.left = left;
                node.right = right;
                stack.Push(node);
            }
            else
            {
                Node node = new Node(c);
                stack.Push(node);
            }
        }
 
        Node root = stack.Peek();
        stack.Pop();
 
        return root;
    }
 
    // Function to print the infix expression for the tree
    static void InOrder(Node node)
    {
        if (node != null)
        {
            InOrder(node.left);
            Console.Write(node.data + " ");
            InOrder(node.right);
        }
    }
 
    // Function to print the postfix expression for the tree
    static void PostOrder(Node node)
    {
        if (node != null)
        {
            PostOrder(node.left);
            PostOrder(node.right);
            Console.Write(node.data + " ");
        }
    }
 
    public static void Main(string[] args)
    {
        string prefix = "*+ab-cd";
        Node root = BuildTree(prefix);
 
        Console.WriteLine("Infix expression is:");
        InOrder(root);
 
        Console.WriteLine("\nPostfix expression is:");
        PostOrder(root);
    }
}

Javascript

// Javascript code to construct an expression tree
// from prefix expression
 
// Represents a node of the required tree
class Node {
constructor(c) {
this.data = c;
this.left = this.right = null;
}
}
 
// Function to recursively build the expression tree
function add(a) {
 
 
// If its the end of the expression
if (a === '') {
    return '';
}
 
// If the character is an operand
if (a.charCodeAt(0) >= 'a'.charCodeAt(0) && a.charCodeAt(0) <= 'z'.charCodeAt(0)) {
    return [new Node(a[0]), a.slice(1)];
} else {
 
    // Create a node with a[0] as the data and
    // both the children set to null
    let p = new Node(a[0]);
     
    // Build the left sub-tree
    let left = add(a.slice(1));
    p.left = left[0];
    let q = left[1];
     
    // Build the right sub-tree
    let right = add(q);
    p.right = right[0];
    q = right[1];
    return [p, q];
}
}
 
// Function to print the infix expression for the tree
function inr(p) { //recursion
 
if (p === null) {
    return;
} else {
    inr(p.left);
    console.log(p.data + ' ');
    inr(p.right);
}
}
 
// Function to print the postfix expression for the tree
function postr(p) {
 
 
if (p === null) {
    return;
} else {
    postr(p.left);
    postr(p.right);
    console.log(p.data + ' ');
}
}
 
// Driver code
 
let a = "*+ab-cd";
let sAndA = add(a);
let s = sAndA[0];
console.log("The Infix expression is:");
inr(s);
console.log("<br>")
console.log("The Postfix expression is:");
postr(s);

Output

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

Time complexity: O(n) because we are scanning all the characters in the given expression
Auxiliary space: O(1)

Convert Ternary Expression to a Binary Tree

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Practice Tags :

Building Expression tree from Prefix Expression

C++

C

Java

Python3

C#

Javascript

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