Data structures 当问题要求返回树时，应返回什么数据结构作为最终实体

因此，我正在研究一个LeetCode问题，从给定的有序和预有序树遍历序列构造一棵树。给定的骨架代码如下所示：

# 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 buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:

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据我所知，我们是否希望在构建树后返回根节点？

是的，我们将返回树的根节点（并直接回答LinkedList类型的问题）：

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以下是LeetCode的官方解决方案，并附有注释：

class Solution:
    def buildTree(self, preorder, inorder):
        """
        :type preorder: List[int]
        :type inorder: List[int]
        :rtype: TreeNode
        """
        def helper(in_left = 0, in_right = len(inorder)):
            nonlocal pre_idx
            # if there is no elements to construct subtrees
            if in_left == in_right:
                return None
            
            # pick up pre_idx element as a root
            root_val = preorder[pre_idx]
            root = TreeNode(root_val)

            # root splits inorder list
            # into left and right subtrees
            index = idx_map[root_val]

            # recursion 
            pre_idx += 1
            # build left subtree
            root.left = helper(in_left, index)
            # build right subtree
            root.right = helper(index + 1, in_right)
            return root
        
        # start from first preorder element
        pre_idx = 0
        # build a hashmap value -> its index
        idx_map = {val:idx for idx, val in enumerate(inorder)} 
        return helper()

class Solution:
    def buildTree(self, preorder, inorder):
        """
        :type preorder: List[int]
        :type inorder: List[int]
        :rtype: TreeNode
        """
        def helper(in_left = 0, in_right = len(inorder)):
            nonlocal pre_idx
            # if there is no elements to construct subtrees
            if in_left == in_right:
                return None
            
            # pick up pre_idx element as a root
            root_val = preorder[pre_idx]
            root = TreeNode(root_val)

            # root splits inorder list
            # into left and right subtrees
            index = idx_map[root_val]

            # recursion 
            pre_idx += 1
            # build left subtree
            root.left = helper(in_left, index)
            # build right subtree
            root.right = helper(index + 1, in_right)
            return root
        
        # start from first preorder element
        pre_idx = 0
        # build a hashmap value -> its index
        idx_map = {val:idx for idx, val in enumerate(inorder)} 
        return helper()