988. Smallest String Starting From Leaf - 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 smallestFromLeaf(self, root: Optional[TreeNode]) -> str:
        def dfs(root, cur):
            if not root:
                return

            cur = chr(ord('a') + root.val) + cur
            if root.left and root.right:
                return min(
                    dfs(root.left, cur),
                    dfs(root.right, cur)
                )

            if root.right:
                return dfs(root.right, cur)
            if root.left:
                return dfs(root.left, cur)
            return cur

        return dfs(root, "")

# 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 smallestFromLeaf(self, root: Optional[TreeNode]) -> str:
        q = deque([(root, "")])
        res = None

        while q:
            node, cur = q.popleft()
            cur = chr(ord('a') + node.val) + cur

            if not node.left and not node.right:
                res = min(res, cur) if res else cur

            if node.left:
                q.append((node.left, cur))
            if node.right:
                q.append((node.right, cur))

        return res

# 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 smallestFromLeaf(self, root: Optional[TreeNode]) -> str:
        stack = [(root, "")]
        res = None

        while stack:
            node, cur = stack.pop()
            cur = chr(ord('a') + node.val) + cur

            if not node.left and not node.right:
                res = min(res, cur) if res else cur

            if node.right:
                stack.append((node.right, cur))
            if node.left:
                stack.append((node.left, cur))

        return res