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. Lets the found index is 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 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); } |
python3
# 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
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