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
SOLUTION:
# 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], isLeft = False, isRight = False) -> str:
if root.left and root.right:
return "{}({})({})".format(root.val, self.tree2str(root.left), self.tree2str(root.right))
if root.left:
return "{}({})".format(root.val, self.tree2str(root.left))
if root.right:
return "{}()({})".format(root.val, self.tree2str(root.right))
return "{}".format(root.val)
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