Convert Ternary Expression to a Binary Tree (original) (raw)
Last Updated : 14 Sep, 2023
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 .
- If we see Symbol '?'
- then we add next character as the left child of root.
- 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 // expression 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 rootFunction 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 // expression 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
JavaScript
`
**Time Complexity : O(n) [ here **n is length of String ]
**Auxiliary Space: **O(n)
Approach for Converting Ternary Expression to Binary Tree.
- The algorithm uses a recursive approach to build the tree in a top-down manner.
- It starts with creating a new node for the current character at the current index.
- If the next character is a '?', it means that the current node needs a left child. So, the algorithm calls itself recursively with the next index to create the left child of the current node.
- If the next character is a ':', it means that the current node needs a right child. So, the algorithm calls itself recursively with the next index to create the right child of the current node.
- Finally, the algorithm returns the root node of the binary tree.
**Here is the implementation of above approach:-
C++ `
#include <bits/stdc++.h> using namespace std;
class Node { public: char val; Node *left, *right; Node(char v) : val(v), left(nullptr), right(nullptr) {} };
Node* ternaryToTree(string exp) { if (exp.empty()) return nullptr; Node *root = new Node(exp[0]); stack<Node*> st; st.push(root); for (int i = 1; i < exp.size(); i += 2) { Node *cur = st.top(); st.pop(); if (exp[i] == '?') { cur->left = new Node(exp[i+1]); st.push(cur); st.push(cur->left); } else { cur->right = new Node(exp[i+1]); st.push(cur->right); } } return root; }
void printTree(Node* root) { if (!root) return; cout << root->val << " "; printTree(root->left); printTree(root->right); }
int main() { string exp = "a?b?c:d:e"; Node *root = ternaryToTree(exp); printTree(root); return 0; }
Java
import java.util.*;
class Node { char val; Node left, right; public Node(char v) { val = v; } }
class Solution { public static Node ternaryToTree(String exp) { if (exp == null || exp.length() == 0) return null; Node root = new Node(exp.charAt(0)); Stack st = new Stack<>(); st.push(root); for (int i = 1; i < exp.length(); i += 2) { Node cur = st.pop(); if (exp.charAt(i) == '?') { cur.left = new Node(exp.charAt(i+1)); st.push(cur); st.push(cur.left); } else { cur.right = new Node(exp.charAt(i+1)); st.push(cur.right); } } return root; }
public static void printTree(Node root) {
if (root == null) return;
System.out.print(root.val + " ");
printTree(root.left);
printTree(root.right);
}
public static void main(String[] args) {
String exp = "a?b?c:d:e";
Node root = ternaryToTree(exp);
printTree(root);
}}
Python3
class Node: def init(self, val): self.val = val self.left = None self.right = None
def ternary_to_tree(exp): if not exp: return None root = Node(exp[0]) stack = [root] for i in range(1, len(exp)): if exp[i] == '?': stack[-1].left = Node(exp[i+1]) stack.append(stack[-1].left) elif exp[i] == ':': stack.pop() while stack[-1].right: stack.pop() stack[-1].right = Node(exp[i+1]) stack.append(stack[-1].right) return root
def print_tree(root): if not root: return print(root.val, end=' ') print_tree(root.left) print_tree(root.right)
if name == "main": exp = "a?b?c:d:e" root = ternary_to_tree(exp) print_tree(root)
C#
using System; using System.Collections.Generic;
class Node{ public char val; public Node left, right; public Node(char v) { val = v; } }
class Solution{ // Function to convert a ternary expression to a tree public static Node ternaryToTree(string exp){ if (exp == null || exp.Length == 0) return null; Node root = new Node(exp[0]); // Create the root node Stack st = new Stack(); // Stack to track nodes st.Push(root); // Push root to the stack for (int i = 1; i < exp.Length; i += 2){ Node cur = st.Pop(); // Pop the top node from the stack if (exp[i] == '?'){ cur.left = new Node(exp[i + 1]); // Create a left child node st.Push(cur); // Push the current node back to the stack st.Push(cur.left); // Push the left child node to the stack } else{ cur.right = new Node(exp[i + 1]); // Create a right child node st.Push(cur.right); // Push the right child node to the stack } } return root; // Return the root node of the tree }
// Function to print the tree in pre-order traversal
public static void printTree(Node root){
if (root == null) return;
Console.Write(root.val + " "); // Print the current node
printTree(root.left); // Recursively print the left subtree
printTree(root.right); // Recursively print the right subtree
}
public static void Main(string[] args){
string exp = "a?b?c:d:e";
Node root = ternaryToTree(exp);
printTree(root);
}}
// THIS CODE IS CONTRIBUTED BY CHANDAN AGARWAL
JavaScript
class Node { constructor(val) { this.val = val; this.left = null; this.right = null; } }
let i = 0;
function convertExpression(expression) { if (!expression || i >= expression.length) { return null; }
const root = new Node(expression[i]); i++;
if (i < expression.length && expression[i] === "?") { i++; root.left = convertExpression(expression); }
if (i < expression.length && expression[i] === ":") { i++; root.right = convertExpression(expression); }
return root; }
function printTree(root) { if (!root) { return; }
console.log(root.val + " "); printTree(root.left); printTree(root.right); }
const stringExpression = "a?b?c:d:e"; const rootNode = convertExpression(stringExpression); printTree(rootNode);
`
**Time complexity: **O(n) - Since we visit each character of the expression exactly once.
**Space complexity: **O(n) - Since in the worst case, the recursion stack can grow to the height of the tree, which can be O(n) if the ternary expression is a degenerate tree (a long chain of '?'). Additionally, we store O(n) nodes in the binary tree.