Construct Binary Tree from String with bracket representation
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 funtcion 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 parenthesisint 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 stringNode* 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 Codeint 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 nodeclass newNode: def __init__(self, data): self.data = data self.left = self.right = None# This funtcion is here just to testdef preOrder(node): if (node == None): return print(node.data, end=" ") preOrder(node.left) preOrder(node.right)# function to return the index of# close parenthesisdef 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 Stringdef 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 Codeif __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 |
Output
4 2 3 1 6 5
Time Complexity: O(N2)
Auxiliary Space: O(N)
Another recursive approach:
Algorithm:
- The very first element of the string is the root.
- 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.
- Once the left child is added recursively, we will look for consecutive “(” and add the right child to the parent node.
- Encountering “)” means the end of either left or right node and we will increment the start index
- The recursion ends when the start index is greater than equal to the end index
Python3
class newNode: def __init__(self, data): self.data = data self.left = self.right = Nonedef 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 returndef treeFromString(string): root = newNode(string[0]) if len(string) > 1: si = [1] ei = len(string)-1 treeFromStringHelper(si, ei, string, root) return root# Driver Codeif __name__ == '__main__': Str = "4(2(3)(1))(6(5))" root = treeFromString(Str) preOrder(root)# This code is contributed by dheerajalimchandani |
Output
4 2 3 1 6 5
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