问题的解决方案超时:从inorder和postorder构建二叉树

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def buildTree(inorder, postorder):
    def helper(in_left, in_right):
        nonlocal post_idx
        if in_left > in_right:
            return None
        
        # 后序遍历的最后一个元素是根节点
        root_val = postorder[post_idx]
        root = TreeNode(root_val)
        post_idx -= 1
        
        # 在中序遍历中找到根节点的位置
        index = idx_map[root_val]
        
        # 先构建右子树，因为后序遍历中根节点前的元素是右子树
        root.right = helper(index + 1, in_right)
        # 再构建左子树
        root.left = helper(in_left, index - 1)
        return root
    
    post_idx = len(postorder) - 1
    idx_map = {val: idx for idx, val in enumerate(inorder)}
    return helper(0, len(inorder) - 1)

# 示例使用
inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
root = buildTree(inorder, postorder)