算法题：字符串构造二叉树

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

def create_tree(root, vals):
    if len(vals) == 0:                       # vals用完了直接返回
        return root
    if vals[0] != '#':                       # 构建root, root.left, root.right
        root = TreeNode(vals[0])
        vals.pop(0)
        root.left = create_tree(root.left, vals)
        root.right = created_tree(root.right, vals)
        return root                          # 本次递归要返回给上一次的本层构造好的树的根节点
    else:
        root = None
        vals.pop(0)
        return root                          # 本次递归要返回给上一次的本层构造好的树的根节点

def print_binary_tree(root, level=0, prefix="Root: "):
    if root:
        print('  ' * level + prefix + str(root.val))
        if root.left or root.right:
            print_binary_tree(root.left, level + 1, "Left: ")
            print_binary_tree(root.right, level + 1, "Right: ")

root = None
strs = "abc##d##e##"
vals = list(strs)
roots = create_tree(root, vals)
print_binary_tree(roots)

不要纠结于递归的整体流程，会让自己迷失，关注本层操作。