Construct String from Binary Tree

lcpy gen -n 606   # by problem number
lcpy gen -s construct_string_from_binary_tree   # by problem name

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Problem

• Node Representation: Each node in the tree should be represented by its integer value.
• Parentheses for Children: If a node has at least one child (either left or right), its children should be represented inside parentheses. Specifically:
  • If a node has a left child, the value of the left child should be enclosed in parentheses immediately following the node’s value.
  • If a node has a right child, the value of the right child should also be enclosed in parentheses. The parentheses for the right child should follow those of the left child.
• Omitting Empty Parentheses: Any empty parentheses pairs (i.e., ()) should be omitted from the final string representation of the tree, with one specific exception: when a node has a right child but no left child. In such cases, you must include an empty pair of parentheses to indicate the absence of the left child. This ensures that the one-to-one mapping between the string representation and the original binary tree structure is maintained.

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Examples

Input: root = [1,2,3,4]
Output: "1(2(4))(3)"
Explanation: Originally, it needs to be "1(2(4)())(3()())", but you need to omit all the empty parenthesis pairs. And it will be "1(2(4))(3)".

Input: root = [1,2,3,null,4]
Output: "1(2()(4))(3)"
Explanation: Almost the same as the first example, except the () after 2 is necessary to indicate the absence of a left child for 2 and the presence of the right child.

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Constraints

• The number of nodes in the tree is in the range [1, 10^4].
• -1000 <= Node.val <= 1000

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Solution

from leetcode_py import TreeNode


class Solution:
    # Time: O(n)
    # Space: O(n)
    def tree2str(self, root: TreeNode[int] | None) -> str:
        if root is None:
            return ""
        left = self.tree2str(root.left)
        right = self.tree2str(root.right)
        if root.left is None and root.right is not None:
            return f"{root.val}()({right})"
        if root.right is not None:
            return f"{root.val}({left})({right})"
        if root.left is not None:
            return f"{root.val}({left})"
        return str(root.val)

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Complexity

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