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Convert Ternary Expression to a Binary Tree

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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;
}


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 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 
// 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




<script>
   
// Javascript program to convert a ternary 
// expreesion to a tree. 
  
// Class to represent Tree node 
class Node
{
    constructor(item)
    {
        this.data = item;
        this.left = null;
        this.right = null;
    }
}
  
// Function to convert Ternary Expression 
// to a Binary Tree. It return the root of tree 
function convertExpression(expression, i)
{
      
    // Base case 
    if (i >= expression.length)
    {
        return null;
    }
      
    // Store current character of 
    // expression_string [ 'a' to 'z'] 
    var 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 
function printTree(root)
{
    if (root == null)
    {
        return;
    }
    document.write(root.data + " ");
    printTree(root.left);
    printTree(root.right);
}
  
// Driver code
var exp = "a?b?c:d:e";
var expression = exp.split('');
var root = convertExpression(expression, 0);
printTree(root);
  
// This code is contributed by noob2000
  
</script>


Output

a b c d e 

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<Node> 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<Node> st = new Stack<Node>(); // 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);


Output

a b c d e 

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.

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Last Updated : 14 Sep, 2023
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