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Convert Ternary Expression to Binary Tree

In this article, we are going to see how to convert the ternary expression to a binary tree? This problem has been featured in Facebook interview.
Submitted by Radib Kar, on March 02, 2019

Problem statement:

Given a string that contains ternary expressions. The expressions may be nested. You need to convert the given ternary expression to a binary Tree and return the root.

Example:

    Input:
    a?b:c

    Output:
      a
     / \
    b   c

    Input:
    a?b?c:d:e

    Output:
        a
       / \
      b   e
     / \
    c   d

Solution:

Ternary expression: In C, we are acquainted with the ternary expression. Ternary expressions are equivalent to the if-else statement in C.

    if(a)
        b
    else
        c

    a,b,c=statements/expressions
    This if else statement is equivalent to a?b:c

Similarly, a ternary expression can be converted to a binary tree, where a will be the root, b will be the left child and c will be the right one. This small miniature can be expanded (followed) for nesting ternary expression.

The algorithm for constructing the binary tree from the ternary expression is:

Pre-requisite:

  1. Ternary expression str
  2. Node* newNode(string str, index i) : Creates a new node with data value str[i]

Algorithm:

    //recursive function to build the tree
    FUNCTION convertExpression(string str, int& i) 
    1.  Create root referring to the current character(str[i]);
    root =newNode(str,i); //create a node with element str[i]
    2.  Increment i (index that point to current character);
        //if i<string length and current token is '?' increment i
    3.  IF(i<str.length() && str[i]=='?'){ 
	    //need to build left child recursively increment i
        root->left=convertExpression(str,i); 
	    //need to build right child recursively
        root->right=convertExpression(str,i); 
    4.  return root;

Algorithm with example:

Tree nodes represented by their values only
Input string (ternary expression)
a?b:c
-------------------------------------------------
In main function we call convertExpression(str,0)
convertExpression(str,0):
root=newNode(str, 0); //root=a
i=1; //incremented
i<str.length() && str[i]=='?'
i=2;//incremented
root->left=convertExpression(str,2);
-------------------------------------------------

convertExpression(str,2):
root=newNode(str, 2); //root=b
i=3; //incremented
i<str.length()&& str[i]!='?'
return root //b
at convertExpression(str,0)
now root->left=b
i=4; //i++ step evaluated
root->right=convertExpression(str,4)
-------------------------------------------------

convertExpression(str,4):
root=newNode(str, 4); //root=c
i=5; //incremented
I is not <str.length()
return root //c
at convertExpression(str,0)
now root->right=c
return root;

returns to main
built tree:
  a
 / \
b   c

C++ implementation

#include <bits/stdc++.h>
using namespace std;

struct Node{
	char data;
	Node *left,*right;
};

//function to create 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 traverse preorder
void preorder(Node *root){ 
	if(root==NULL)
		return;
	cout<<root->data<<" ";
	preorder(root->left);
	preorder(root->right);
}

//recursive function to build the tree
Node *convertExpression(string str,int& i) 
{
	Node* root=newNode(str[i]);
	i++;
	if(i<str.length() && str[i]=='?'){
		i++;
		root->left=convertExpression(str,i);
		i++; //skipping ':' character
		root->right=convertExpression(str,i);
	}
	return root;
}


int main(){
	string str;
	
	cout<<"Enter your expression\n";
	cin>>str;
	
	int i=0;
	Node *root=convertExpression(str,i);
	cout<<"Printing pre-order traversal of the tree...\n";
	//pre-order traversal of the tree, 
	//should be same with expressionthe 
	preorder(root);
	cout<<endl;
	
	return 0;
}

Output

First run:
Enter your expression
a?b:c
Printing pre-order traversal of the tree...
a b c

Second run:
Enter your expression
a?b?c:e:d
Printing pre-order traversal of the tree...
a b c e d





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