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606 Construct String From Binary Tree - Easy

Problem:

Given the root of a binary tree, construct a string consisting of parenthesis and integers from a binary tree with the preorder traversal way, and return it.

Omit all the empty parenthesis pairs that do not affect the one-to-one mapping relationship between the string and the original binary tree.

Example 1:

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 unnecessary empty parenthesis pairs. And it will be "1(2(4))(3)"

Example 2:

Input: root = [1,2,3,null,4] Output: "1(2()(4))(3)" Explanation: Almost the same as the first example, except we cannot omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • -1000 <= Node.val <= 1000

Problem Analysis:

  • recursively traverse the tree in preorder manner

  • If logic:

    • if left exists:
      • we want to wrap left_node.val with parenthesis
    • if right exists:
      • we want to wrap right_node.val with parenthesis
    • if left don't exist but right exists (Edge case, case 2):
      • we want an empty parenthesis as placeholder for left node
      • we want to wrap right_node.val with parenthesis

  • Simplifying the logic, we just need:

    • if left_str or right_str:
    • if right_str:

  • Time Complexity:

    • The time complexity remains O(n), where n is the number of nodes in the binary tree, as each node is visited once.

  • Space Complexity:

    • The space complexity is O(h), where h is the height of the binary tree, due to the recursive call stack.

Solutions:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def tree2str(self, root: Optional[TreeNode]) -> str:
        if not root:
            return ""

        left_str = self.tree2str(root.left)
        right_str = self.tree2str(root.right)

        result = str(root.val)

        if left_str or right_str:
            result += "(" + left_str + ")"

        if right_str:
            result += "(" + right_str + ")"

        return result

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