Skip to content
Related Articles

Related Articles

Improve Article

Construct Binary Tree from String with bracket representation

  • Difficulty Level : Medium
  • Last Updated : 12 Jul, 2021

Construct a binary tree from a string consisting of parenthesis and integers. The whole input represents a binary tree. It contains an integer followed by zero, one or two pairs of parenthesis. The integer represents the root’s value and a pair of parenthesis contains a child binary tree with the same structure. Always start to construct the left child node of the parent first if it exists.

Examples: 

Input : "1(2)(3)" 
Output : 1 2 3
Explanation :
           1
          / \
         2   3
Explanation: first pair of parenthesis contains 
left subtree and second one contains the right 
subtree. Preorder of above tree is "1 2 3".  

Input : "4(2(3)(1))(6(5))"
Output : 4 2 3 1 6 5
Explanation :
           4
         /   \
        2     6
       / \   / 
      3   1 5   

We know first character in string is root. Substring inside the first adjacent pair of parenthesis is for left subtree and substring inside second pair of parenthesis is for right subtree as in the below diagram. 
 

We need to find the substring corresponding to left subtree and substring corresponding to right subtree and then recursively call on both of the substrings. 



For this first find the index of starting index and end index of each substring. 
To find the index of closing parenthesis of left subtree substring, use a stack. Let the found index be stored in index variable. 

C++




/* C++ program to construct a binary tree from
   the given string */
#include <bits/stdc++.h>
using namespace std;
 
/* A binary tree node has data, pointer to left
   child and a pointer to right child */
struct Node {
    int data;
    Node *left, *right;
};
/* Helper function that allocates a new node */
Node* newNode(int data)
{
    Node* node = (Node*)malloc(sizeof(Node));
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}
 
/* This function is here just to test  */
void preOrder(Node* node)
{
    if (node == NULL)
        return;
    printf("%d ", node->data);
    preOrder(node->left);
    preOrder(node->right);
}
 
// function to return the index of close parenthesis
int findIndex(string str, int si, int ei)
{
    if (si > ei)
        return -1;
 
    // Inbuilt stack
    stack<char> s;
 
    for (int i = si; i <= ei; i++) {
 
        // if open parenthesis, push it
        if (str[i] == '(')
            s.push(str[i]);
 
        // if close parenthesis
        else if (str[i] == ')') {
            if (s.top() == '(') {
                s.pop();
 
                // if stack is empty, this is
                // the required index
                if (s.empty())
                    return i;
            }
        }
    }
    // if not found return -1
    return -1;
}
 
// function to construct tree from string
Node* treeFromString(string str, int si, int ei)
{
    // Base case
    if (si > ei)
        return NULL;
 
    // new root
    Node* root = newNode(str[si] - '0');
    int index = -1;
 
    // if next char is '(' find the index of
    // its complement ')'
    if (si + 1 <= ei && str[si + 1] == '(')
        index = findIndex(str, si + 1, ei);
 
    // if index found
    if (index != -1) {
 
        // call for left subtree
        root->left = treeFromString(str, si + 2, index - 1);
 
        // call for right subtree
        root->right
            = treeFromString(str, index + 2, ei - 1);
    }
    return root;
}
 
// Driver Code
int main()
{
    string str = "4(2(3)(1))(6(5))";
    Node* root = treeFromString(str, 0, str.length() - 1);
    preOrder(root);
}

Java




/* Java program to cona binary tree from
   the given String */
import java.util.*;
class GFG
{
 
  /* A binary tree node has data, pointer to left
   child and a pointer to right child */
  static class Node
  {
    int data;
    Node left, right;
  };
 
  /* Helper function that allocates a new node */
  static Node newNode(int data)
  {
    Node node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
  }
 
  /* This funtcion is here just to test  */
  static void preOrder(Node node)
  {
    if (node == null)
      return;
    System.out.printf("%d ", node.data);
    preOrder(node.left);
    preOrder(node.right);
  }
 
  // function to return the index of close parenthesis
  static int findIndex(String str, int si, int ei)
  {
    if (si > ei)
      return -1;
 
    // Inbuilt stack
    Stack<Character> s = new Stack<>();
    for (int i = si; i <= ei; i++)
    {
 
      // if open parenthesis, push it
      if (str.charAt(i) == '(')
        s.add(str.charAt(i));
 
      // if close parenthesis
      else if (str.charAt(i) == ')')
      {
        if (s.peek() == '(')
        {
          s.pop();
 
          // if stack is empty, this is
          // the required index
          if (s.isEmpty())
            return i;
        }
      }
    }
 
    // if not found return -1
    return -1;
  }
 
  // function to contree from String
  static Node treeFromString(String str, int si, int ei)
  {
 
    // Base case
    if (si > ei)
      return null;
 
    // new root
    Node root = newNode(str.charAt(si) - '0');
    int index = -1;
 
    // if next char is '(' find the index of
    // its complement ')'
    if (si + 1 <= ei && str.charAt(si+1) == '(')
      index = findIndex(str, si + 1, ei);
 
    // if index found
    if (index != -1)
    {
 
      // call for left subtree
      root.left = treeFromString(str, si + 2, index - 1);
 
      // call for right subtree
      root.right
        = treeFromString(str, index + 2, ei - 1);
    }
    return root;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    String str = "4(2(3)(1))(6(5))";
    Node root = treeFromString(str, 0, str.length() - 1);
    preOrder(root);
  }
}
 
// This code is contributed by gauravrajput1

Python




# Python3 program to conStruct a
# binary tree from the given String
 
# Helper class that allocates a new node
 
 
class newNode:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None
 
# This funtcion is here just to test
 
 
def preOrder(node):
    if (node == None):
        return
    print(node.data, end=" ")
    preOrder(node.left)
    preOrder(node.right)
 
# function to return the index of
# close parenthesis
 
 
def findIndex(Str, si, ei):
    if (si > ei):
        return -1
 
    # Inbuilt stack
    s = []
    for i in range(si, ei + 1):
 
        # if open parenthesis, push it
        if (Str[i] == '('):
            s.append(Str[i])
 
        # if close parenthesis
        elif (Str[i] == ')'):
            if (s[-1] == '('):
                s.pop(-1)
 
                # if stack is empty, this is
                # the required index
                if len(s) == 0:
                    return i
    # if not found return -1
    return -1
 
# function to conStruct tree from String
 
 
def treeFromString(Str, si, ei):
 
    # Base case
    if (si > ei):
        return None
 
    # new root
    root = newNode(ord(Str[si]) - ord('0'))
    index = -1
 
    # if next char is '(' find the
    # index of its complement ')'
    if (si + 1 <= ei and Str[si + 1] == '('):
        index = findIndex(Str, si + 1, ei)
 
    # if index found
    if (index != -1):
 
        # call for left subtree
        root.left = treeFromString(Str, si + 2,
                                   index - 1)
 
        # call for right subtree
        root.right = treeFromString(Str, index + 2,
                                    ei - 1)
    return root
 
 
# Driver Code
if __name__ == '__main__':
    Str = "4(2(3)(1))(6(5))"
    root = treeFromString(Str, 0, len(Str) - 1)
    preOrder(root)
 
# This code is contributed by pranchalK

C#




/* C# program to cona binary tree from
   the given String */
using System;
using System.Collections.Generic;
 
public class GFG
{
 
  /* A binary tree node has data, pointer to left
   child and a pointer to right child */
  public
 
 class Node
  {
    public
 
 int data;
    public
 
 Node left, right;
  };
 
  /* Helper function that allocates a new node */
  static Node newNode(int data)
  {
    Node node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
  }
 
  /* This funtcion is here just to test  */
  static void preOrder(Node node)
  {
    if (node == null)
      return;
    Console.Write("{0} ", node.data);
    preOrder(node.left);
    preOrder(node.right);
  }
 
  // function to return the index of close parenthesis
  static int findIndex(String str, int si, int ei)
  {
    if (si > ei)
      return -1;
 
    // Inbuilt stack
    Stack<char> s = new Stack<char>();
    for (int i = si; i <= ei; i++)
    {
 
      // if open parenthesis, push it
      if (str[i] == '(')
        s.Push(str[i]);
 
      // if close parenthesis
      else if (str[i] == ')')
      {
        if (s.Peek() == '(')
        {
          s.Pop();
 
          // if stack is empty, this is
          // the required index
          if (s.Count==0)
            return i;
        }
      }
    }
 
    // if not found return -1
    return -1;
  }
 
  // function to contree from String
  static Node treeFromString(String str, int si, int ei)
  {
 
    // Base case
    if (si > ei)
      return null;
 
    // new root
    Node root = newNode(str[si] - '0');
    int index = -1;
 
    // if next char is '(' find the index of
    // its complement ')'
    if (si + 1 <= ei && str[si+1] == '(')
      index = findIndex(str, si + 1, ei);
 
    // if index found
    if (index != -1)
    {
 
      // call for left subtree
      root.left = treeFromString(str, si + 2, index - 1);
 
      // call for right subtree
      root.right
        = treeFromString(str, index + 2, ei - 1);
    }
    return root;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    String str = "4(2(3)(1))(6(5))";
    Node root = treeFromString(str, 0, str.Length - 1);
    preOrder(root);
  }
}
 
// This code is contributed by gauravrajput1

Javascript




<script>
/* Javascript program to cona binary tree from
   the given String */
 
 /* A binary tree node has data, pointer to left
   child and a pointer to right child */
class Node
{
    constructor()
    {
        this.data = 0;
        this.left = this.right = null;
    }
}
 
/* Helper function that allocates a new node */
function newNode(data)
{
    let node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
}
 
/* This funtcion is here just to test  */
function preOrder(node)
{
    if (node == null)
      return;
    document.write(node.data + " ");
    preOrder(node.left);
    preOrder(node.right);
}
 
  // function to return the index of close parenthesis
function findIndex(str, si, ei)
{
    if (si > ei)
      return -1;
  
    // Inbuilt stack
    let s = [];
    for (let i = si; i <= ei; i++)
    {
  
      // if open parenthesis, push it
      if (str[i] == '(')
        s.push(str[i]);
  
      // if close parenthesis
      else if (str[i] == ')')
      {
        if (s[s.length-1] == '(')
        {
          s.pop();
  
          // if stack is empty, this is
          // the required index
          if (s.length == 0)
            return i;
        }
      }
    }
  
    // if not found return -1
    return -1;
}
 
// function to contree from String
function treeFromString(str,si,ei)
{
    // Base case
    if (si > ei)
      return null;
  
    // new root
    let root = newNode(str[si].charCodeAt(0) - '0'.charCodeAt(0));
    let index = -1;
  
    // if next char is '(' find the index of
    // its complement ')'
    if (si + 1 <= ei && str[si + 1] == '(')
      index = findIndex(str, si + 1, ei);
  
    // if index found
    if (index != -1)
    {
  
      // call for left subtree
      root.left = treeFromString(str, si + 2, index - 1);
  
      // call for right subtree
      root.right
        = treeFromString(str, index + 2, ei - 1);
    }
    return root;
}
 
 // Driver Code
let str = "4(2(3)(1))(6(5))";
let root = treeFromString(str, 0, str.length - 1);
preOrder(root);
 
// This code is contributed by patel2127
</script>
Output
4 2 3 1 6 5 

Time Complexity: O(N2)
Auxiliary Space: O(N)

Another recursive approach:

Algorithm:

  1. The very first element of the string is the root.
  2. If the next two consecutive elements are “(” and “)”, this means there is no left child otherwise we will create and add the left child to the parent node recursively.
  3. Once the left child is added recursively, we will look for consecutive “(” and add the right child to the parent node.
  4. Encountering “)” means the end of either left or right node and we will increment the start index
  5. The recursion ends when the start index is greater than equal to the end index

C++




#include <bits/stdc++.h>
using namespace std;
 
// custom data type for tree building
struct Node {
    int data;
    struct Node* left;
    struct Node* right;
    Node(int val)
    {
        data = val;
        left = right = NULL;
    }
};
 
// Below function accepts sttring and a pointer variable as
// an argument
// and draw the tree. Returns the root of the tree
Node* constructtree(string s, int* start)
{
    // Assuming there is/are no negative
    // character/characters in the string
    if (s.size() == 0 || *start >= s.size())
        return NULL;
 
    // constructing a number from the continuous digits
    int num = 0;
    while (*start < s.size() && s[*start] != '('
           && s[*start] != ')') {
        int num_here = (int)(s[*start] - '0');
        num = num * 10 + num_here;
        *start = *start + 1;
    }
 
    // creating a node from the constructed number from
    // above loop
    struct Node* root = new Node(num);
 
    // check if start has reached the end of the string
    if (*start >= s.size())
        return root;
 
    // As soon as we see first right parenthesis from the
    // current node we start to construct the tree in the
    // left
    if (*start < s.size() && s[*start] == '(') {
        *start = *start + 1;
        root->left = constructtree(s, start);
    }
    if (*start < s.size() && s[*start] == ')')
        *start = *start + 1;
 
    // As soon as we see second right parenthesis from the
    // current node we start to construct the tree in the
    // right
    if (*start < s.size() && s[*start] == '(') {
        *start = *start + 1;
        root->right = constructtree(s, start);
    }
    if (*start < s.size() && s[*start] == ')')
        *start = *start + 1;
    return root;
}
void preorder(Node* root)
{
    if (root == NULL)
        return;
    cout << root->data << " ";
    preorder(root->left);
    preorder(root->right);
}
int main()
{
    string s = "4(2(3)(1))(6(5))";
    // cin>>s;
    int start = 0;
    Node* root = constructtree(s, &start);
    preorder(root);
    return 0;
}
//This code is contributed by Chaitanya Sharma.

Java




import java.io.*;
import java.util.*;
 
class GFG{
     
// Node class for the Tree
static class Node
{
    int data;
    Node left,right;
     
    Node(int data)
    {
        this.data = data;
        this.left = this.right = null;
    }
}
 
// static variable to point to the
// starting index of the string.
static int start = 0;
 
// Construct Tree Function which accepts
// a string and return root of the tree;
static Node constructTree(String s)
{
     
    // Check for null or empty string
    // and return null;
    if (s.length() == 0 || s == null)
    {
        return null;
    }
     
    if (start >= s.length())
        return null;
     
    // Boolean variable to check
    // for negative numbers
    boolean neg = false;
     
    // Condition to check for negative number
    if (s.charAt(start) == '-')
    {
        neg = true;
        start++;
    }
     
    // This loop basically construct the
    // number from the continous digits
    int num = 0;
    while (start < s.length() &&
           Character.isDigit(s.charAt(start)))
    {
        int digit = Character.getNumericValue(
            s.charAt(start));
        num = num * 10 + digit;
        start++;
    }
     
    // If string contains - minus sign
    // then append - to the number;
    if (neg)
        num = -num;
     
    // Create the node object i.e. root of
    // the tree with data = num;
    Node node = new Node(num);
     
    if (start >= s.length())
    {
        return node;
    }
     
    // Check for open bracket and add the
    // data to the left subtree recursively
    if (start < s.length() && s.charAt(start) == '(' )
    {
        start++;
        node.left = constructTree(s);
    }
     
    if (start < s.length() && s.charAt(start) == ')')
    {
        start++;
        return node;
    }
     
    // Check for open bracket and add the data
    // to the right subtree recursively
    if (start < s.length() && s.charAt(start) == '(')
    {
        start++;
        node.right = constructTree(s);
    }
     
    if (start < s.length() && s.charAt(start) == ')')
    {
        start++;
        return node;
    }
    return node;
}
 
// Print tree function
public static void printTree(Node node)
{
    if (node == null)
        return;
   
    System.out.println(node.data + " ");
    printTree(node.left);
    printTree(node.right);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Input
    String s = "4(2(3)(1))(6(5))";
   
    // Call the function cunstruct tree
    // to create the tree pass the string;
    Node root = constructTree(s);
   
    // Function to print preorder of the tree
    printTree(root);
}
}
 
// This code is contributed by yash181999

Python3




class newNode:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None
 
 
def preOrder(node):
    if (node == None):
        return
    print(node.data, end=" ")
    preOrder(node.left)
    preOrder(node.right)
 
 
def treeFromStringHelper(si, ei, arr, root):
 
    if si[0] >= ei:
        return None
 
    if arr[si[0]] == "(":
 
        if arr[si[0]+1] != ")":
            if root.left is None:
                if si[0] >= ei:
                    return
                new_root = newNode(arr[si[0]+1])
                root.left = new_root
                si[0] += 2
                treeFromStringHelper(si, ei, arr, new_root)
 
        else:
            si[0] += 2
 
        if root.right is None:
            if si[0] >= ei:
                return
 
            if arr[si[0]] != "(":
                si[0] += 1
                return
 
            new_root = newNode(arr[si[0]+1])
            root.right = new_root
            si[0] += 2
            treeFromStringHelper(si, ei, arr, new_root)
        else:
            return
 
    if arr[si[0]] == ")":
        if si[0] >= ei:
            return
        si[0] += 1
        return
 
    return
 
 
def treeFromString(string):
 
    root = newNode(string[0])
 
    if len(string) > 1:
        si = [1]
        ei = len(string)-1
 
        treeFromStringHelper(si, ei, string, root)
 
    return root
 
# Driver Code
if __name__ == '__main__':
    Str = "4(2(3)(1))(6(5))"
    root = treeFromString(Str)
    preOrder(root)
 
# This code is contributed by dheerajalimchandani

Javascript




<script>
 
// Node class for the Tree
class Node
{
    constructor(data)
    {
        this.data=data;
        this.left = this.right = null;
    }
}
 
// static variable to point to the
// starting index of the string.
let start = 0;
 
// Construct Tree Function which accepts
// a string and return root of the tree;
function constructTree(s)
{
    // Check for null or empty string
    // and return null;
    if (s.length == 0 || s == null)
    {
        return null;
    }
      
    if (start >= s.length)
        return null;
      
    // Boolean variable to check
    // for negative numbers
    let neg = false;
      
    // Condition to check for negative number
    if (s[start] == '-')
    {
        neg = true;
        start++;
    }
      
    // This loop basically construct the
    // number from the continous digits
    let num = 0;
    while (start < s.length && !isNaN(s[start] -
    parseInt(s[start])))
    {
        let digit = parseInt(
            s[start]);
        num = num * 10 + digit;
        start++;
    }
      
    // If string contains - minus sign
    // then append - to the number;
    if (neg)
        num = -num;
      
    // Create the node object i.e. root of
    // the tree with data = num;
    let node = new Node(num);
      
    if (start >= s.length)
    {
        return node;
    }
      
    // Check for open bracket and add the
    // data to the left subtree recursively
    if (start < s.length && s[start] == '(' )
    {
        start++;
        node.left = constructTree(s);
    }
      
    if (start < s.length && s[start] == ')')
    {
        start++;
        return node;
    }
      
    // Check for open bracket and add the data
    // to the right subtree recursively
    if (start < s.length && s[start] == '(')
    {
        start++;
        node.right = constructTree(s);
    }
      
    if (start < s.length && s[start] == ')')
    {
        start++;
        return node;
    }
    return node;
}
 
// Print tree function
function printTree(node)
{
    if (node == null)
        return;
    
    document.write(node.data + " ");
    printTree(node.left);
    printTree(node.right);
}
 
// Driver Code
// Input
let s = "4(2(3)(1))(6(5))";
 
// Call the function cunstruct tree
// to create the tree pass the string;
let root = constructTree(s);
 
// Function to print preorder of the tree
printTree(root);
 
// This code is contributed by unknown2108
 
</script>
Output
4 2 3 1 6 5 

This article is contributed by Chhavi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :