Construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way.
The null node needs to be represented by empty parenthesis pair “()”. Omit all the empty parenthesis pairs that don’t affect the one-to-one mapping relationship between the string and the original binary tree.
Examples:
Input : Preorder: [1, 2, 3, 4]
1
/ \
2 3
/
4
Output: "1(2(4))(3)"
Explanation: Originally it needs to be "1(2(4)
())(3()())", but we need to omit all the
unnecessary empty parenthesis pairs.
And it will be "1(2(4))(3)".
Input : Preorder: [1, 2, 3, null, 4]
1
/ \
2 3
\
4
Output: "1(2()(4))(3)"
This is opposite of Construct Binary Tree from String with bracket representation
The idea is to do the preorder traversal of the given Binary Tree along with this, we need to make use of braces at appropriate positions. But, we also need to make sure that we omit the unnecessary braces. We print the current node and call the same given function for the left and the right children of the node in that order(if they exist). For every node encountered, the following cases are possible.
Case 1: Both the left child and the right child exist for the current node. In this case, we need to put the braces () around both the left child’s preorder traversal output and the right child’s preorder traversal output.
Case 2: None of the left or the right child exist for the current node. In this case, as shown in the figure below, considering empty braces for the null left and right children is redundant. Hence, we need not put braces for any of them.
Case 3: Only the left child exists for the current node. As the figure below shows, putting empty braces for the right child in this case is unnecessary while considering the preorder traversal. This is because the right child will always come after the left child in the preorder traversal. Thus, omitting the empty braces for the right child also leads to same mapping between the string and the binary tree.
Case 4: Only the right child exists for the current node. In this case, we need to consider the empty braces for the left child. This is because, during the preorder traversal, the left child needs to be considered first. Thus, to indicate that the child following the current node is a right child we need to put a pair of empty braces for the left child.
C++
/* C++ program to construct string from binary tree*/#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);} // Function to construct string from binary treevoid treeToString(Node* root, string& str){ // bases case if (root == NULL) return; // push the root data as character str.push_back(root->data + '0'); // if leaf node, then return if (!root->left && !root->right) return; // for left subtree str.push_back('('); treeToString(root->left, str); str.push_back(')'); // only if right child is present to // avoid extra parenthesis if (root->right) { str.push_back('('); treeToString(root->right, str); str.push_back(')'); }} // Driver Codeint main(){ /* Let us construct below tree 1 / \ 2 3 / \ \ 4 5 6 */ struct Node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->right = newNode(6); string str = ""; treeToString(root, str); cout << str;} |
Java
// Java program to construct string from binary tree 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; }; static String str; /* 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); } // Function to construct string from binary tree static void treeToString(Node root) { // bases case if (root == null) return; // push the root data as character str += (Character.valueOf((char) (root.data + '0'))); // if leaf node, then return if (root.left == null && root.right == null) return; // for left subtree str += ('('); treeToString(root.left); str += (')'); // only if right child is present to // avoid extra parenthesis if (root.right != null) { str += ('('); treeToString(root.right); str += (')'); } } // Driver Code public static void main(String[] args) { /* Let us construct below tree 1 / \ 2 3 / \ \ 4 5 6 */ Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.right = newNode(6); str = ""; treeToString(root); System.out.println(str);}} // This code is contributed by 29AjayKumar |
Python3
# Python3 program to construct string from binary tree # A binary tree node has data, pointer to left# child and a pointer to right childclass Node: def __init__(self, data): self.data = data self.left = None self.right = None # Function to construct string from binary treedef treeToString(root: Node, string: list): # base case if root is None: return # push the root data as character string.append(str(root.data)) # if leaf node, then return if not root.left and not root.right: return # for left subtree string.append('(') treeToString(root.left, string) string.append(')') # only if right child is present to # avoid extra parenthesis if root.right: string.append('(') treeToString(root.right, string) string.append(')') # Driver Codeif __name__ == "__main__": # Let us construct below tree # 1 # / \ # 2 3 # / \ \ # 4 5 6 root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) root.right.right = Node(6) string = [] treeToString(root, string) print(''.join(string)) # This code is contributed by# sanjeev2552 |
C#
// C# program to construct string from binary tree using System; 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; }; static String str; /* 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); } // Function to construct string from binary tree static void treeToString(Node root) { // bases case if (root == null) return; // push the root data as character str += (char)(root.data + '0'); // if leaf node, then return if (root.left == null && root.right == null) return; // for left subtree str += ('('); treeToString(root.left); str += (')'); // only if right child is present to // avoid extra parenthesis if (root.right != null) { str += ('('); treeToString(root.right); str += (')'); } } // Driver Code public static void Main(String[] args) { /* Let us construct below tree 1 / \ 2 3 / \ \ 4 5 6 */ Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.right = newNode(6); str = ""; treeToString(root); Console.WriteLine(str);}} // This code is contributed by Princi Singh |
Output:
1(2(4)(5))(3()(6))
Time complexity : O(n) The preorder traversal is done over the n nodes.
Space complexity : O(n). The depth of the recursion tree can go upto n in case of a skewed tree.
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