Convert Ternary Expression to a Binary Tree

Given a string that contains ternary expressions. The expressions may be nested, task is convert the given ternary expression to a binary Tree.

Examples:

Input :  string expression =   a?b:c 
Output :        a
              /  \
             b    c

Input : expression =  a?b?c:d:e
Output :     a
           /  \
          b    e
        /  \
       c    d

Asked In : Facebook Interview

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Idea is that we traverse a string make first character as root and do following step recursively .
1. If we see Symbol ‘?’
…….. then we add next character as the left child of root.
2. If we see Symbol ‘:’
…….. then we add it as the right child of current root.
do this process until we traverse all element of “String”.

Below is the implementation of above idea

C++

// C++ program to convert a ternary expression to 
// a tree. 
#include<bits/stdc++.h> 
using namespace std; 
  
// tree structure 
struct Node 
{ 
    char data; 
    Node *left, *right; 
}; 
  
// function create a new node 
Node *newNode(char Data) 
{ 
    Node *new_node = new Node; 
    new_node->data = Data; 
    new_node->left = new_node->right = NULL; 
    return new_node; 
} 
  
// Function to convert Ternary Expression to a Binary 
// Tree. It return the root of tree 
//Notice that we pass index i by reference because we want to skip the characters in the subtree 
Node *convertExpression(string str, int & i) 
{ 
    // store current character of expression_string  
    // [ 'a' to 'z']  
    Node * root =newNode(str[i]); 
  
    //If it was last character return 
    //Base Case 
    if(i==str.length()-1) return root; 
  
    // Move ahead in str  
    i++; 
    //If the next character is '?'.Then there will be subtree for the current node 
    if(str[i]=='?') 
    { 
        //skip the '?' 
        i++; 
  
        //construct the left subtree 
        //Notice after the below recursive call i will point to ':' just before the right child of current node since we pass i by reference 
        root->left = convertExpression(str,i); 
          
        //skip the ':' character 
        i++; 
  
        //construct the right subtree 
        root->right = convertExpression(str,i); 
        return root; 
    } 
    //If the next character is not '?' no subtree just return it 
    else return root; 
} 
  
// function print tree 
void printTree( Node *root) 
{ 
    if (!root) 
        return ; 
    cout << root->data <<" "; 
    printTree(root->left); 
    printTree(root->right); 
} 
  
// Driver program to test above function 
int main() 
{ 
    string expression = "a?b?c:d:e"; 
    int i=0; 
    Node *root = convertExpression(expression, i); 
    printTree(root) ; 
    return 0; 
} 

Java

// Java program to convert a ternary  
// expreesion to a tree. 
import java.util.Queue; 
import java.util.LinkedList; 
   
// Class to represent Tree node  
class Node  
{ 
    char data; 
    Node left, right; 
   
    public Node(char item)  
    { 
        data = item; 
        left = null; 
        right = null; 
    } 
} 
   
// Class to convert a ternary expression to a Tree  
class BinaryTree  
{ 
    // Function to convert Ternary Expression to a Binary 
    // Tree. It return the root of tree 
    Node convertExpression(char[] expression, int i) 
    { 
        // Base case 
        if (i >= expression.length) 
            return null; 
       
        // store current character of expression_string 
        // [ 'a' to 'z'] 
        Node root = new Node(expression[i]); 
       
        // Move ahead in str 
        ++i; 
       
        // if current character of ternary expression is '?' 
        // then we add next character as a left child of 
        // current node 
        if (i < expression.length && expression[i]=='?') 
            root.left = convertExpression(expression, i+1); 
       
        // else we have to add it as a right child of 
        // current node expression.at(0) == ':' 
        else if (i < expression.length) 
            root.right = convertExpression(expression, i+1); 
       
        return root; 
    } 
      
    // function print tree 
    public void printTree( Node root) 
    { 
        if (root == null) 
            return; 
                  
        System.out.print(root.data +" "); 
        printTree(root.left); 
        printTree(root.right); 
    } 
      
// Driver program to test above function 
    public static void main(String args[])  
    { 
        String exp = "a?b?c:d:e"; 
        BinaryTree tree = new BinaryTree(); 
        char[] expression=exp.toCharArray();  
        Node root = tree.convertExpression(expression, 0); 
        tree.printTree(root) ; 
    } 
} 
  
/* This code is contributed by Mr. Somesh Awasthi */

Python3

# Class to define a node  
# structure of the tree 
class Node: 
    def __init__(self, key): 
        self.data = key 
        self.left = None
        self.right = None
  
# Function to convert ternary  
# expression to a Binary tree 
# It returns the root node  
# of the tree 
def convert_expression(expression, i): 
    if i >= len(expression): 
        return None
  
    # Create a new node object 
    # for the expression at 
    # ith index 
    root = Node(expression[i]) 
  
    i += 1
  
    # if current character of  
    # ternary expression is '?' 
    # then we add next character  
    # as a left child of 
    # current node 
    if (i < len(expression) and 
                expression[i] is "?"): 
        root.left = convert_expression(expression, i + 1) 
          
    # else we have to add it  
    # as a right child of 
    # current node expression[0] == ':' 
    elif i < len(expression): 
        root.right = convert_expression(expression, i + 1) 
    return root 
  
# Function to print the tree 
# in a pre-order traversal pattern 
def print_tree(root): 
    if not root: 
        return
    print(root.data, end=' ') 
    print_tree(root.left) 
    print_tree(root.right) 
  
# Driver Code 
if __name__ == "__main__": 
    string_expression = "a?b?c:d:e"
    root_node = convert_expression(string_expression, 0) 
    print_tree(root_node) 
  
# This code is contributed 
# by Kanav Malhotra 

C#

// C# program to convert a ternary  
// expreesion to a tree.  
using System; 
  
// Class to represent Tree node  
public class Node 
{ 
    public char data; 
    public Node left, right; 
  
    public Node(char item) 
    { 
        data = item; 
        left = null; 
        right = null; 
    } 
} 
  
// Class to convert a ternary  
// expression to a Tree  
public class BinaryTree 
{ 
    // Function to convert Ternary Expression  
    // to a Binary Tree. It return the root of tree  
    public virtual Node convertExpression(char[] expression, 
                                          int i) 
    { 
        // Base case  
        if (i >= expression.Length) 
        { 
            return null; 
        } 
  
        // store current character of  
        // expression_string [ 'a' to 'z']  
        Node root = new Node(expression[i]); 
  
        // Move ahead in str  
        ++i; 
  
        // if current character of ternary expression  
        // is '?' then we add next character as a  
        // left child of current node  
        if (i < expression.Length && expression[i] == '?') 
        { 
            root.left = convertExpression(expression, i + 1); 
        } 
  
        // else we have to add it as a right child  
        // of current node expression.at(0) == ':'  
        else if (i < expression.Length) 
        { 
            root.right = convertExpression(expression, i + 1); 
        } 
  
        return root; 
    } 
  
    // function print tree  
    public virtual void printTree(Node root) 
    { 
        if (root == null) 
        { 
            return; 
        } 
  
        Console.Write(root.data + " "); 
        printTree(root.left); 
        printTree(root.right); 
    } 
  
// Driver Code 
public static void Main(string[] args) 
{ 
    string exp = "a?b?c:d:e"; 
    BinaryTree tree = new BinaryTree(); 
    char[] expression = exp.ToCharArray(); 
    Node root = tree.convertExpression(expression, 0); 
    tree.printTree(root); 
} 
} 
  
// This code is contributed by Shrikant13 

Output :

a b c d e

Time Complexity : O(n) [ here n is length of String ]

This article is contributed by Nishant Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

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