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
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; } |
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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 */ |
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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 |
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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 |
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Output :
a b c d e
Time Complexity : O(n) [ here n is length of String ]
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