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Construct Binary Tree from String with bracket representation
• Difficulty Level : Medium
• Last Updated : 20 Nov, 2020

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 ` `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 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);` `}`

## 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`

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

## 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`

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 contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.