Python 将可视化的二叉树放入元组

发布于10月01日

我得到了这棵二叉树

我想把它放在一个具有以下 struct 的元组中(Left_Subtree，key，right_subtree)(其中Left_Subtree和Right_Subtree本身就是元组. 我怎样才能以正确的方式做这件事呢？

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


tree_tuple = (((None, 6, None), 4,2,(9, 7, 10)), 1,
              ((13, 11, 14),8, (15, 12, 16), 5, 3))


print(len(tree_tuple))


def parse_tuple(data):
    if isinstance(data, tuple) and len(data) == 3:
        node = TreeNode(data[1])
        node.left = parse_tuple(data[0])
        node.right = parse_tuple(data[2])
    elif data is None:
        node = None
    else:
        node = TreeNode(data)
    return node


tree = parse_tuple(tree_tuple)

print(tree.right.right.left.right.key)

我试过了，但显然我的元组 struct 甚至我的树本身有问题

推荐答案

我想这就是你想要做的.

您的parse_tuple确实也给出了12作为答案，问题出在您的tree_tuple的格式中.

我只是修改了一下我将如何处理它.

下一步是不考虑任何问题:)

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


def parse_tuple(data):
    if isinstance(data, tuple) and len(data) == 3:
        node = TreeNode(data[1])
        node.left = parse_tuple(data[0])
        node.right = parse_tuple(data[2])
        return node
    elif data is None:
        return None
    else:
        raise ValueError("Invalid Tuple Format")


tree_tuple = (
    (((None, 6, None), 4, ((None, 9, None), 7, (None, 10, None))), 2, None),
    1,
    (
        None,
        3,
        (
            (
                ((None, 13, None), 11, (None, 14, None)),
                8,
                ((None, 15, None), 12, (None, 16, None)),
            ),
            5,
            None,
        ),
    ),
)

tree = parse_tuple(tree_tuple)


print("LEFT")
print(tree.left.key)
print(tree.left.left.key)
print(tree.left.left.left.key)
print(tree.left.left.right.key)
print(tree.left.left.right.left.key)
print(tree.left.left.right.right.key)

print("\nRIGHT")
print(tree.right.key)
print(tree.right.right.key)
print(tree.right.right.left.key)
print(tree.right.right.left.left.key)
print(tree.right.right.left.left.left.key)
print(tree.right.right.left.left.right.key)
print(tree.right.right.left.right.key)
print(tree.right.right.left.right.left.key)
print(tree.right.right.left.right.right.key)