606. Construct String From Binary Tree - Explanation

# 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 ""

        cur = root.val
        left = self.tree2str(root.left)
        right = self.tree2str(root.right)

        if left and right:
            return f"{cur}({left})({right})"

        if right:
            return f"{cur}()({right})"

        if left:
            return f"{cur}({left})"

        return str(cur)

# 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:
        res = []

        def preorder(root):
            if not root:
                return
            res.append("(")
            res.append(str(root.val))
            if not root.left and root.right:
                res.append("()")
            preorder(root.left)
            preorder(root.right)
            res.append(")")

        preorder(root)
        return "".join(res)[1:-1]

# 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 ""

        res = []
        stack = []
        last_visited = None
        cur = root

        while cur or stack:
            if cur:
                res.append(f"({cur.val}")
                if not cur.left and cur.right:
                    res.append("()")

                stack.append(cur)
                cur = cur.left
            else:
                top = stack[-1]
                if top.right and last_visited != top.right:
                    cur = top.right
                else:
                    stack.pop()
                    res.append(")")
                    last_visited = top

        return "".join(res)[1:-1]