# 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;

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:
a?b:c
Printing pre-order traversal of the tree...
a b c

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