Skip to content
Related Articles

Related Articles

Improve Article

Program to convert Infix notation to Expression Tree

  • Difficulty Level : Medium
  • Last Updated : 27 Aug, 2021

Given a string representing infix notation. The task is to convert it to an expression tree.
Expression Tree is a binary tree where the operands are represented by leaf nodes and operators are represented by intermediate nodes. No node can have a single child.

Construction of Expression tree

The algorithm follows a combination of shunting yard along with postfix-to-expression tree conversion.

Consider the below line:

((s[i]!='^' && p[stC.top()]>=p[s[i]]) || 
(s[i]=='^' && p[stC.top()]>p[s[i]])))

You might remember that unlike ‘+’, ‘-‘, ‘*’ and ‘/’; ‘^’ is right associative. 
In simpler terms, a^b^c is a^(b^c) not (a^b)^c. So it must be evaluated from the right.



Now lets take a look at how does the algorithm work,(Take a quick glance at the code to get a better idea of the variables used) 

Let us have an expression s = ((a+b)*c-e*f)
currently both the stacks are empty- 
(we'll use C to denote the char stack and N for node stack)
    s[0] = '('            ((a+b)*c-e*f)
                          ^
        C|(|, N| |
    
    s[1] = '('           ((a+b)*c-e*f)
                          ^
         |(|  
        C|(|, N| |
                         
    s[2] = 'a'            ((a+b)*c-e*f)
                            ^
         |(|  
        C|(|, N|a|
                         
    s[3] = '+'            ((a+b)*c-e*f)
                             ^
         |+|
         |(|  
        C|(|, N|a|
        
    s[4] = 'b'             ((a+b)*c-e*f)
                               ^
         |+|
         |(|   |b|
        C|(|, N|a|
         
    s[5] = ')'             ((a+b)*c-e*f)
                                ^
         |+|            t = '+'         +
         |(|   |b|  ->  t1= 'b'        / \    ->  
        C|(|, N|a|      t2= 'a'       a   b      C|(|, N|+|
        
    s[6] = '*'              ((a+b)*c-e*f)
                                  ^
         |*|  
        C|(|, N|+|
    
    s[7] = 'c'              ((a+b)*c-e*f)
                                   ^
         |*|   |c|
        C|(|, N|+|
        
    s[8] = '-'   ((a+b)*c-e*f)              now (C.top(*)>s[8](-))
                         ^    t = '*'        *
         |*|   |c|            t1 = c        / \  ->    |-|
        C|(|, N|+|            t2 = +       +   c      C|(|, N|*|
                                          / \
                                         a   b
    s[9] = 'e'            ((a+b)*c-e*f)
                                   ^
         |-|   |e|
        C|(|, N|*|
        
    s[10] = '*'            ((a+b)*c-e*f)      now (C.top(-)>s[10](*))
                                     ^
         |*|                          
         |-|   |e| 
        C|(|, N|*|
    
    s[11] = 'f'             ((a+b)*c-e*f)
                                       ^
         |*|   |f|
         |-|   |e|
        C|(|, N|*|
    
    s[12] = ')'             ((a+b)*c-e*f)
       1>                               ^
         |*|   |f|         t = '*'          *        
         |-|   |e|  ->     t1= 'f'  ->     / \  ->   |-|   |*|
        C|(|, N|*|         t2= 'e'        e   f     C|(|, N|*|
        
       2>                               
                           t = '-'           -        
         |-|   |*|  ->     t1= '*'  ->     /   \  ->   
        C|(|, N|*|         t2= '*'        *     *     C| |, N|-|
                                         / \   / \
                                        +   c  e  f
                                       / \
                                      a   b 
      now make (-) the root of the tree

Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Tree Structure
typedef struct node
{
    char data;
    struct node *left, *right;
} * nptr;
 
// Function to create new node
nptr newNode(char c)
{
    nptr n = new node;
    n->data = c;
    n->left = n->right = nullptr;
    return n;
}
 
// Function to build Expression Tree
nptr build(string& s)
{
 
    // Stack to hold nodes
    stack<nptr> stN;
 
    // Stack to hold chars
    stack<char> stC;
    nptr t, t1, t2;
 
    // Prioritising the operators
    int p[123] = { 0 };
    p['+'] = p['-'] = 1, p['/'] = p['*'] = 2, p['^'] = 3,
    p[')'] = 0;
 
    for (int i = 0; i < s.length(); i++)
    {
        if (s[i] == '(') {
 
            // Push '(' in char stack
            stC.push(s[i]);
        }
 
        // Push the operands in node stack
        else if (isalpha(s[i]))
        {
            t = newNode(s[i]);
            stN.push(t);
        }
        else if (p[s[i]] > 0)
        {
            // If an operator with lower or
            // same associativity appears
            while (
                !stC.empty() && stC.top() != '('
                && ((s[i] != '^' && p[stC.top()] >= p[s[i]])
                    || (s[i] == '^'
                        && p[stC.top()] > p[s[i]])))
            {
 
                // Get and remove the top element
                // from the character stack
                t = newNode(stC.top());
                stC.pop();
 
                // Get and remove the top element
                // from the node stack
                t1 = stN.top();
                stN.pop();
 
                // Get and remove the currently top
                // element from the node stack
                t2 = stN.top();
                stN.pop();
 
                // Update the tree
                t->left = t2;
                t->right = t1;
 
                // Push the node to the node stack
                stN.push(t);
            }
 
            // Push s[i] to char stack
            stC.push(s[i]);
        }
        else if (s[i] == ')') {
            while (!stC.empty() && stC.top() != '(')
            {
                t = newNode(stC.top());
                stC.pop();
                t1 = stN.top();
                stN.pop();
                t2 = stN.top();
                stN.pop();
                t->left = t2;
                t->right = t1;
                stN.push(t);
            }
            stC.pop();
        }
    }
    t = stN.top();
    return t;
}
 
// Function to print the post order
// traversal of the tree
void postorder(nptr root)
{
    if (root)
    {
        postorder(root->left);
        postorder(root->right);
        cout << root->data;
    }
}
 
// Driver code
int main()
{
    string s = "(a^b^(c/d/e-f)^(x*y-m*n))";
    s = "(" + s;
    s += ")";
    nptr root = build(s);
   
    // Function call
    postorder(root);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Tree Structure
static class nptr
{
    char data;
    nptr left, right;
} ;
 
// Function to create new node
static nptr newNode(char c)
{
    nptr n = new nptr();
    n.data = c;
    n.left = n.right = null;
    return n;
}
 
// Function to build Expression Tree
static nptr build(String s)
{
 
    // Stack to hold nodes
    Stack<nptr> stN = new Stack<>();
 
    // Stack to hold chars
    Stack<Character> stC = new Stack<>();
    nptr t, t1, t2;
 
    // Prioritising the operators
    int []p = new int[123];
    p['+'] = p['-'] = 1;
    p['/'] = p['*'] = 2;
    p['^'] = 3;
    p[')'] = 0;
 
    for (int i = 0; i < s.length(); i++)
    {
        if (s.charAt(i) == '(') {
 
            // Push '(' in char stack
            stC.add(s.charAt(i));
        }
 
        // Push the operands in node stack
        else if (Character.isAlphabetic(s.charAt(i)))
        {
            t = newNode(s.charAt(i));
            stN.add(t);
        }
        else if (p[s.charAt(i)] > 0)
        {
           
            // If an operator with lower or
            // same associativity appears
            while (
                !stC.isEmpty() && stC.peek() != '('
                && ((s.charAt(i) != '^' && p[stC.peek()] >= p[s.charAt(i)])
                    || (s.charAt(i) == '^'
                        && p[stC.peek()] > p[s.charAt(i)])))
            {
 
                // Get and remove the top element
                // from the character stack
                t = newNode(stC.peek());
                stC.pop();
 
                // Get and remove the top element
                // from the node stack
                t1 = stN.peek();
                stN.pop();
 
                // Get and remove the currently top
                // element from the node stack
                t2 = stN.peek();
                stN.pop();
 
                // Update the tree
                t.left = t2;
                t.right = t1;
 
                // Push the node to the node stack
                stN.add(t);
            }
 
            // Push s[i] to char stack
            stC.push(s.charAt(i));
        }
        else if (s.charAt(i) == ')') {
            while (!stC.isEmpty() && stC.peek() != '(')
            {
                t = newNode(stC.peek());
                stC.pop();
                t1 = stN.peek();
                stN.pop();
                t2 = stN.peek();
                stN.pop();
                t.left = t2;
                t.right = t1;
                stN.add(t);
            }
            stC.pop();
        }
    }
    t = stN.peek();
    return t;
}
 
// Function to print the post order
// traversal of the tree
static void postorder(nptr root)
{
    if (root != null)
    {
        postorder(root.left);
        postorder(root.right);
        System.out.print(root.data);
    }
}
 
// Driver code
public static void main(String[] args)
{
    String s = "(a^b^(c/d/e-f)^(x*y-m*n))";
    s = "(" + s;
    s += ")";
    nptr root = build(s);
   
    // Function call
    postorder(root);
}
}
 
// This code is contributed by aashish1995

Python3




# Python3 implementation of the approach
 
# Tree Structure
class nptr:
   
    # Constructor to set the data of
    # the newly created tree node
    def __init__(self, c):
        self.data = c
        self.left = None
        self.right = None
 
# Function to create new node
def newNode(c):
    n = nptr(c)
    return n
 
# Function to build Expression Tree
def build(s):
   
    # Stack to hold nodes
    stN = []
 
    # Stack to hold chars
    stC = []
 
    # Prioritising the operators
    p = [0]*(123)
    p[ord('+')] = p[ord('-')] = 1
    p[ord('/')] = p[ord('*')] = 2
    p[ord('^')] = 3
    p[ord(')')] = 0
 
    for i in range(len(s)):
        if (s[i] == '('):
            # Push '(' in char stack
            stC.append(s[i])
 
        # Push the operands in node stack
        elif (s[i].isalpha()):
            t = newNode(s[i])
            stN.append(t)
        elif (p[ord(s[i])] > 0):
           
            # If an operator with lower or
            # same associativity appears
            while (len(stC) != 0 and stC[-1] != '(' and ((s[i] != '^' and p[ord(stC[-1])] >= p[ord(s[i])])
                    or (s[i] == '^'and
                    p[ord(stC[-1])] > p[ord(s[i])]))):
               
                # Get and remove the top element
                # from the character stack
                t = newNode(stC[-1])
                stC.pop()
 
                # Get and remove the top element
                # from the node stack
                t1 = stN[-1]
                stN.pop()
 
                # Get and remove the currently top
                # element from the node stack
                t2 = stN[-1]
                stN.pop()
 
                # Update the tree
                t.left = t2
                t.right = t1
 
                # Push the node to the node stack
                stN.append(t)
 
            # Push s[i] to char stack
            stC.append(s[i])
             
        elif (s[i] == ')'):
            while (len(stC) != 0 and stC[-1] != '('):
                t = newNode(stC[-1])
                stC.pop()
                t1 = stN[-1]
                stN.pop()
                t2 = stN[-1]
                stN.pop()
                t.left = t2
                t.right = t1
                stN.append(t)
            stC.pop()
    t = stN[-1]
    return t
 
# Function to print the post order
# traversal of the tree
def postorder(root):
    if (root != None):
        postorder(root.left)
        postorder(root.right)
        print(root.data, end = "")
 
s = "(a^b^(c/d/e-f)^(x*y-m*n))"
s = "(" + s
s += ")"
root = build(s)
 
# Function call
postorder(root)
 
# This code is contributed by divyeshrabadiya07.

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
 
// Tree Structure
public  class nptr
{
   public char data;
   public nptr left, right;
} ;
 
// Function to create new node
static nptr newNode(char c)
{
    nptr n = new nptr();
    n.data = c;
    n.left = n.right = null;
    return n;
}
 
// Function to build Expression Tree
static nptr build(String s)
{
 
    // Stack to hold nodes
    Stack<nptr> stN = new Stack<nptr>();
 
    // Stack to hold chars
    Stack<char> stC = new Stack<char>();
    nptr t, t1, t2;
 
    // Prioritising the operators
    int []p = new int[123];
    p['+'] = p['-'] = 1;
    p['/'] = p['*'] = 2;
    p['^'] = 3;
    p[')'] = 0;
 
    for (int i = 0; i < s.Length; i++)
    {
        if (s[i] == '(')
        {
 
            // Push '(' in char stack
            stC.Push(s[i]);
        }
 
        // Push the operands in node stack
        else if (char.IsLetter(s[i]))
        {
            t = newNode(s[i]);
            stN.Push(t);
        }
        else if (p[s[i]] > 0)
        {
           
            // If an operator with lower or
            // same associativity appears
            while (stC.Count != 0 && stC.Peek() != '('
                && ((s[i] != '^' && p[stC.Peek()] >= p[s[i]])
                    || (s[i] == '^'&& p[stC.Peek()] > p[s[i]])))
            {
 
                // Get and remove the top element
                // from the character stack
                t = newNode(stC.Peek());
                stC.Pop();
 
                // Get and remove the top element
                // from the node stack
                t1 = stN.Peek();
                stN.Pop();
 
                // Get and remove the currently top
                // element from the node stack
                t2 = stN.Peek();
                stN.Pop();
 
                // Update the tree
                t.left = t2;
                t.right = t1;
 
                // Push the node to the node stack
                stN.Push(t);
            }
 
            // Push s[i] to char stack
            stC.Push(s[i]);
        }
        else if (s[i] == ')')
        {
            while (stC.Count != 0 && stC.Peek() != '(')
            {
                t = newNode(stC.Peek());
                stC.Pop();
                t1 = stN.Peek();
                stN.Pop();
                t2 = stN.Peek();
                stN.Pop();
                t.left = t2;
                t.right = t1;
                stN.Push(t);
            }
            stC.Pop();
        }
    }
    t = stN.Peek();
    return t;
}
 
// Function to print the post order
// traversal of the tree
static void postorder(nptr root)
{
    if (root != null)
    {
        postorder(root.left);
        postorder(root.right);
        Console.Write(root.data);
    }
}
 
// Driver code
public static void Main(String[] args)
{
    String s = "(a^b^(c/d/e-f)^(x*y-m*n))";
    s = "(" + s;
    s += ")";
    nptr root = build(s);
   
    // Function call
    postorder(root);
}
}
 
// This code is contributed by aashish1995

Javascript




<script>
 
    // JavaScript implementation of the approach
     
    // Tree Structure
    class nptr
    {
       constructor(c) {
           this.left = null;
           this.right = null;
           this.data = c;
        }
    }
 
    // Function to create new node
    function newNode(c)
    {
        let n = new nptr(c);
        return n;
    }
 
    // Function to build Expression Tree
    function build(s)
    {
 
        // Stack to hold nodes
        let stN = [];
 
        // Stack to hold chars
        let stC = [];
        let t, t1, t2;
 
        // Prioritising the operators
        let p = new Array(123);
        p['+'.charCodeAt()] = p['-'.charCodeAt()] = 1;
        p['/'.charCodeAt()] = p['*'.charCodeAt()] = 2;
        p['^'.charCodeAt()] = 3;
        p[')'.charCodeAt()] = 0;
 
        for (let i = 0; i < s.length; i++)
        {
            if (s[i] == '(')
            {
 
                // Push '(' in char stack
                stC.push(s[i]);
            }
 
            // Push the operands in node stack
            else if ((/[a-zA-Z]/).test(s[i]))
            {
                t = newNode(s[i]);
                stN.push(t);
            }
            else if (p[s[i].charCodeAt()] > 0)
            {
 
                // If an operator with lower or
                // same associativity appears
                while (stC.length != 0 && stC[stC.length - 1] != '('
                    && ((s[i] != '^' &&
                    p[stC[stC.length - 1].charCodeAt()] >=
                    p[s[i].charCodeAt()])
                        || (s[i] == '^'&&
                        p[stC[stC.length - 1].charCodeAt()] >
                        p[s[i].charCodeAt()])))
                {
 
                    // Get and remove the top element
                    // from the character stack
                    t = newNode(stC[stC.length - 1]);
                    stC.pop();
 
                    // Get and remove the top element
                    // from the node stack
                    t1 = stN[stN.length - 1];
                    stN.pop();
 
                    // Get and remove the currently top
                    // element from the node stack
                    t2 = stN[stN.length - 1];
                    stN.pop();
 
                    // Update the tree
                    t.left = t2;
                    t.right = t1;
 
                    // Push the node to the node stack
                    stN.push(t);
                }
 
                // Push s[i] to char stack
                stC.push(s[i]);
            }
            else if (s[i] == ')')
            {
                while (stC.length != 0 &&
                stC[stC.length - 1] != '(')
                {
                    t = newNode(stC[stC.length - 1]);
                    stC.pop();
                    t1 = stN[stN.length - 1];
                    stN.pop();
                    t2 = stN[stN.length - 1];
                    stN.pop();
                    t.left = t2;
                    t.right = t1;
                    stN.push(t);
                }
                stC.pop();
            }
        }
        t = stN[stN.length - 1];
        return t;
    }
 
    // Function to print the post order
    // traversal of the tree
    function postorder(root)
    {
        if (root != null)
        {
            postorder(root.left);
            postorder(root.right);
            document.write(root.data);
        }
    }
     
    let s = "(a^b^(c/d/e-f)^(x*y-m*n))";
    s = "(" + s;
    s += ")";
    let root = build(s);
    
    // Function call
    postorder(root);
     
</script>

 
 

Output
abcd/e/f-xy*mn*-^^^

The time Complexity is O(n) as each character is accessed only once. 
The space Complexity is O(n) as (char_stack + node_stack) <= n
  

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :