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 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; int num = 0; // In case the number is having more than 1 digit while(si <= ei && str[si] >= '0' && str[si] <= '9') { num *= 10; num += (str[si] - '0'); si++; } // new root Node* root = newNode(num); int index = -1; // if next char is '(' find the index of // its complement ')' if (si <= ei && str[si] == '(') index = findIndex(str, si, ei); // if index found if (index != -1) { // call for left subtree root->left = treeFromString(str, si + 1, 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 construct a 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 function 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 construct tree from String static Node treeFromString(String str, int si, int ei) { // Base case if (si > ei) return null; int num = 0; // In case the number is having more than 1 digit while(si <= ei && str.charAt(si) >= '0' && str.charAt(si) <= '9') { num *= 10; num += (str.charAt(si) - '0'); si++; } si--; // new root Node root = newNode(num); 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 function 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 construct a 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 function 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 construct tree from String static Node treeFromString(String str, int si, int ei) { // Base case if (si > ei) return null; int num = 0; // In case the number is having more than 1 digit while(si <= ei && str[si] >= '0' && str[si] <= '9') { num *= 10; num += (str[si] - '0'); si++; } si--; // new root Node root = newNode(num); 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
/* Javascript program to construct a 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 function is here just to test */function preOrder(node){ if (node == null) return; console.log(node.data + " "); preOrder(node.left); preOrder(node.right);} // function to return the index of close parenthesisfunction 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 construct tree from Stringfunction treeFromString(str,si,ei){ // Base case if (si > ei) return null; let num = 0; // In case the number is having more than 1 digit while(si <= ei && str[si] >= '0' && str[si] <= '9') { num *= 10; num += (str[si] - '0'); si++; } si--; // new root let root = newNode(num); 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 Codelet str = "14(2(3)(1))(6(5))";let root = treeFromString(str, 0, str.length - 1);preOrder(root);// This code is contributed by patel2127 |
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
C++
#include <bits/stdc++.h>using namespace std;// custom data type for tree buildingstruct Node { int data; struct Node* left; struct Node* right; Node(int val) { data = val; left = right = NULL; }};// Below function accepts string and a pointer variable as// an argument// and draw the tree. Returns the root of the treeNode* 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 = NULL; if(num > 0) root = new Node(num); // 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; return root; } // 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 Treestatic 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 continuous 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 functionpublic static void printTree(Node node){ if (node == null) return; System.out.println(node.data + " "); printTree(node.left); printTree(node.right);}// Driver Codepublic static void main(String[] args){ // Input String s = "4(2(3)(1))(6(5))"; // Call the function construct 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 = 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 |
C#
using System;class GFG { // Class containing left and // right child of current // node and key value class Node { public int data; public Node left, right; public Node(int data) { this.data = data; left = 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 bool neg = false; // Condition to check for negative number if (s[start] == '-') { neg = true; start++; } // This loop basically construct the // number from the continuous digits int num = 0; while (start < s.Length && Char.IsDigit(s[start])) { int digit = (int)Char.GetNumericValue( 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; 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[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 static void printTree(Node node) { if (node == null) return; Console.Write(node.data + " "); printTree(node.left); printTree(node.right); } // Driver code static void Main() { // Input string s = "4(2(3)(1))(6(5))"; // Call the function construct 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 decode2207. |
Javascript
<script>// Node class for the Treeclass 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 continuous 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 functionfunction printTree(node){ if (node == null) return; document.write(node.data + " "); printTree(node.left); printTree(node.right);}// Driver Code// Inputlet s = "4(2(3)(1))(6(5))";// Call the function construct tree// to create the tree pass the string;let root = constructTree(s);// Function to print preorder of the treeprintTree(root);// This code is contributed by unknown2108</script> |
Output
4 2 3 1 6 5
Time complexity: O(n) where n is the length of the string.
Auxiliary space: O(n).
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