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Expression Tree

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  • Difficulty Level : Medium
  • Last Updated : 23 Jan, 2023
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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:

expressiontre

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. 

  1. If a character is an operand push that into the stack
  2. 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.

Examples:  

Input:  A B C*+ D/
Output: A + B * C / D

The first three symbols are operands, so create tree nodes and push pointers to them onto a stack as shown below.

In the Next step, an operator ‘*’ will going read, so two pointers to trees are popped, a new tree is formed and a pointer to it is pushed onto the stack

In the Next step,  an operator ‘+’ will read, so two pointers to trees are popped, a new tree is formed and a pointer to it is pushed onto the stack.

A3f.png

Similarly, as above cases first we push ‘D’ into the stack and then in the last step first, will read ‘/’ and then as previous step topmost element will pop out and then will be right subtree of root  ‘/’ and other nodes will be right subtree.        

Final Constructed Expression Tree is:

A4f.png

Below is the code of the above approach:  

Below is the implementation of the above approach:  

C++




// C++ program for expression tree
#include <bits/stdc++.h>
using namespace std;
class node {
public:
    char value;
    node* left;
    node* right;
    node* next = NULL;
    node(char c)
    {
        this->value = c;
        left = NULL;
        right = NULL;
    }
    node()
    {
        left = NULL;
        right = NULL;
    }
    friend class Stack;
    friend class expression_tree;
};
class Stack {
    node* head = NULL;
 
public:
    void push(node*);
    node* pop();
    friend class expression_tree;
};
class expression_tree {
public:
    void inorder(node* x)
    {
        // cout<<"Tree in InOrder Traversal is: "<<endl;
        if (x == NULL)
            return;
        else {
            inorder(x->left);
            cout << x->value << "  ";
            inorder(x->right);
        }
    }
};
 
void Stack::push(node* x)
{
    if (head == NULL) {
        head = x;
    }
    // We are inserting here nodes at the top of the stack [following LIFO principle]
    else {
        x->next = head;
        head = x;
    }
}
node* Stack::pop()
{
    // Popping out the top most[ pointed with head] element
    node* p = head;
    head = head->next;
    return p;
}
int main()
{
    string s = "ABC*+D/";
    // If you  wish take input from user:
    //cout << "Insert Postorder Expression: " << endl;
    //cin >> s;
    Stack e;
    expression_tree a;
    node *x, *y, *z;
    int l = s.length();
    for (int i = 0; i < l; i++) {
        // if read character is operator then popping two
        // other elements from stack and making a binary
        // tree
        if (s[i] == '+' || s[i] == '-' || s[i] == '*'
            || s[i] == '/' || s[i] == '^') {
            z = new node(s[i]);
            x = e.pop();
            y = e.pop();
            z->left = y;
            z->right = x;
            e.push(z);
        }
        else {
            z = new node(s[i]);
            e.push(z);
        }
    }
    cout << " The Inorder Traversal of Expression Tree: ";
    a.inorder(z);
    return 0;
}

C




#include <stdio.h>
#include <stdlib.h>
 
/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct node {
    char data;
    struct node* left;
    struct node* right;
    struct node* next;
};
 struct node *head=NULL;
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct node* newNode(char data)
{
    struct node* node
        = (struct node*)malloc(sizeof(struct node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
    node->next = NULL;
     
    return (node);
}
void printInorder(struct node* node)
{
    if (node == NULL)
        return;
    else{
    /* first recur on left child */
    printInorder(node->left);
 
    /* then print the data of node */
    printf("%c ", node->data);
 
    /* now recur on right child */
    printInorder(node->right);
    }
}
 
void push(struct node* x)
{
    if(head==NULL)
    head = x;
    else
    {
        (x)->next = head;
        head  = x;
    }
    // struct node* temp;
    // while(temp!=NULL)
    // {
    //     printf("%c ", temp->data);
    //     temp = temp->next;
    // }
}
struct node* pop()
{
    // Popping out the top most[ pointed with head] element
    struct node* p = head;
    head = head->next;
    return p;
}
int main()
{
    char s[] = {'A','B','C','*','+','D','/'};
    int l = sizeof(s) / sizeof(s[0]) ;
    struct node *x, *y, *z;
    for (int i = 0; i < l; i++) {
        // if read character is operator then popping two
        // other elements from stack and making a binary
        // tree
        if (s[i] == '+' || s[i] == '-' || s[i] == '*'
            || s[i] == '/' || s[i] == '^') {
            z = newNode(s[i]);
            x = pop();
            y = pop();
            z->left = y;
            z->right = x;
            push(z);
        }
        else {
            z = newNode(s[i]);
            push(z);
        }
    }
    printf(" The Inorder Traversal of Expression Tree: ");
    printInorder(z);
    return 0;
}

Java




import java.util.Stack;
 
class Node{
    char data;
    Node left,right;
    public Node(char data){
        this.data = data;
        left = right = null;
    }
}
 
public class Main {
    public static boolean isOperator(char ch){
        if(ch=='+' || ch=='-'|| ch=='*' || ch=='/' || ch=='^'){
            return true;
        }
        return false;
    }
    public static Node expressionTree(String postfix){
        Stack<Node> st = new Stack<Node>();
        Node t1,t2,temp;
 
        for(int i=0;i<postfix.length();i++){
            if(!isOperator(postfix.charAt(i))){
                temp = new Node(postfix.charAt(i));
                st.push(temp);
            }
            else{
                temp = new Node(postfix.charAt(i));
 
                t1 = st.pop();
                t2 = st.pop();
 
                temp.left = t2;
                temp.right = t1;
 
                st.push(temp);
            }
 
        }
        temp = st.pop();
        return temp;
    }
    public static void inorder(Node root){
        if(root==null) return;
 
        inorder(root.left);
        System.out.print(root.data);
        inorder(root.right);
    }
    public static void main(String[] args) {
        String postfix = "ABC*+D/";
 
        Node r = expressionTree(postfix);
        inorder(r);
    }
}

Python3




class Node:
    def __init__(self, value=None, left=None, right=None, next=None):
        self.value = value
        self.left = left
        self.right = right
        self.next = next
 
class Stack:
    def __init__(self):
        self.head = None
 
    def push(self, node):
        if not self.head:
            self.head = node
        else:
            node.next = self.head
            self.head = node
 
    def pop(self):
        if self.head:
            popped = self.head
            self.head = self.head.next
            return popped
        else:
            raise Exception("Stack is empty")
 
class ExpressionTree:
    def inorder(self, x):
        if not x:
            return
        self.inorder(x.left)
        print(x.value, end=" ")
        self.inorder(x.right)
 
def main():
    s = "ABC*+D/"
    stack = Stack()
    tree = ExpressionTree()
    for c in s:
        if c in "+-*/^":
            z = Node(c)
            x = stack.pop()
            y = stack.pop()
            z.left = y
            z.right = x
            stack.push(z)
        else:
            stack.push(Node(c))
    print("The Inorder Traversal of Expression Tree: ", end="")
    tree.inorder(stack.pop())
 
if __name__ == "__main__":
    main()

C#




using System;
using System.Collections.Generic;
 
 
class Node{
    public char data;
    public Node left,right;
    public Node(char data){
        this.data = data;
        left = right = null;
    }
}
 
public class GFG {
    public static bool isOperator(char ch){
        if(ch=='+' || ch=='-'|| ch=='*' || ch=='/' || ch=='^'){
            return true;
        }
        return false;
    }
    static Node expressionTree(String postfix){
        Stack<Node> st = new Stack<Node>();
        Node t1, t2, temp;
 
        for(int i = 0; i < postfix.Length; i++)
        {
            if(!isOperator(postfix[i])){
                temp = new Node(postfix[i]);
                st.Push(temp);
            }
            else{
                temp = new Node(postfix[i]);
 
                t1 = st.Pop();
                t2 = st.Pop();
 
                temp.left = t2;
                temp.right = t1;
 
                st.Push(temp);
            }
 
        }
        temp = st.Pop();
        return temp;
    }
    static void inorder(Node root){
        if(root == null) return;
 
        inorder(root.left);
        Console.Write(root.data);
        inorder(root.right);
    }
    public static void Main(String[] args)
    {
        String postfix = "ABC*+D/";
 
        Node r = expressionTree(postfix);
        inorder(r);
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




// Javascript code for the above approach
class Node {
    constructor(value = null, left = null, right = null, next = null) {
        this.value = value;
        this.left = left;
        this.right = right;
        this.next = next;
    }
}
 
class Stack {
    constructor() {
        this.head = null;
    }
 
    push(node) {
        if (!this.head) {
            this.head = node;
        }
        else
        {
         // We are inserting here nodes at
         // the top of the stack [following LIFO principle]
            node.next = this.head;
            this.head = node;
        }
    }
 
    pop() {
        if (this.head) {
            let popped = this.head;
            this.head = this.head.next;
            return popped;
        } else {
            throw new Error("Stack is empty");
        }
    }
}
 
class ExpressionTree {
    inorder(x) {
        if (!x) {
            return;
        }
        this.inorder(x.left);
        console.log(x.value+" ");
        this.inorder(x.right);
    }
}
 
 
    let s = "ABC*+D/";
    let stack = new Stack();
    let tree = new ExpressionTree();
    for (let c of s) {
        if (c === "+" || c === "-" || c === "*" || c === "/" || c === "^") {
            let z = new Node(c);
            let x = stack.pop();
            let y = stack.pop();
            z.left = y;
            z.right = x;
            stack.push(z);
        } else {
            stack.push(new Node(c));
        }
    }
    console.log("The Inorder Traversal of Expression Tree: ");
    tree.inorder(stack.pop());
 
// This code is contributed by lokeshpotta20.

Output

 The Inorder Traversal of Expression Tree: A  +  B  *  C  /  D  

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