从叶结点开始的最小字符串

这个题解采用了DFS（深度优先搜索）的方法来遍历整个二叉树。在遍历的过程中，使用一个列表 `path` 来存储从根节点到当前节点的路径。当到达一个叶节点（即没有子节点的节点）时，将 `path` 中的字符反转并连接成字符串，以得到从叶节点到根节点的字符串。然后，比较当前生成的字符串与先前找到的最小字符串，如果当前的更小，则更新结果字符串。这个过程持续进行，直到所有的叶节点都被访问过。最后，返回字典序最小的字符串。

时间复杂度: O(n * h)

空间复杂度: O(h) or O(n)

# 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:
        ord_a = ord('a')  # ASCII value of 'a'

        path = []  # This will store the current path from root to the current node
        min_dict_order_str = ""  # This will store the smallest lexicographical string found
        def __traverse(root):
            nonlocal min_dict_order_str
            if not root: return
            alp_char = chr(root.val + ord_a)  # Convert the node value to corresponding character
            path.append(alp_char)
            if not (root.left or root.right):  # Check if it is a leaf node
                curr_str = ''.join(path[::-1])  # Reverse the path to form the required string
                if not min_dict_order_str or curr_str < min_dict_order_str:  # Compare with the smallest string found so far
                    min_dict_order_str = curr_str
            else:
                __traverse(root.left)
                __traverse(root.right)
            path.pop()  # Backtrack
        
        __traverse(root)
        return min_dict_order_str