python 中树的左右旋转

Question

我使用 class:

class Node:
    def __init__(self, value):
        self.key = value
        self.left = None
        self.right = None
        self.parent = None

我创建了这棵树：

n_12 = Node(12)
n_15 = Node(15)
n_3 = Node(3)
n_7 = Node(7)
n_1 = Node(1)
n_2 = Node(2)
n_not1 = Node(-1)

n_12.right = n_15
n_12.left = n_3
n_3.right = n_7
n_3.left = n_1
n_1.right = n_2
n_1.left = n_not1

n_12.parent = None
n_15.parent = n_12
n_3.parent = n_12
n_7.parent = n_3
n_1.parent = n_3
n_2.parent = n_1
n_not1.parent = n_1

我试过这段代码：

def rightRotate(t): 
    if t == None or t.left == None:
        return None
    n = t
    l = t.left
    r = t.right
    lr = t.left.right
    ll = t.left.left
    t = t.left
    t.right = n
    if r != None:
        t.right.right = r
    if lr != None:
        t.right.left = lr
    if ll != None:
        t.left = ll

但它没有用，使用根节点 n_12 它删除了一些节点。为什么它不起作用，我不明白为什么我没有所有节点。如果我调用 rightRotate(n_1)，我有一个无限循环。

Answer 1

你写 “我有一个无限循环”，但你的代码没有循环，所以这一定发生在你代码的其他地方。

我看到两个问题：

1) 赋值应该是无条件的

if lr != None:
    t.right.left = lr

在 lr is None 时也需要此作业。否则，t.right.left 将保持等于 l，即那一刻的 t，因此您确实在树中留下了一个循环。

2) 双线程

您的树是双线程的，即它也有 parent 个链接。但是这些不会在您的 rightRotate 函数中更新。因此，要么不使用 parent 链接（这是更可取的），要么调整您的代码，以便 parent 链接也根据轮换更新。

其他备注：

可以简化以下代码：

if r != None:
    t.right.right = r   # was already equal to r
if lr != None:
    t.right.left = lr   # see above. should not be behind a condition
if ll != None:
    t.left = ll         # was already equal to ll

这样可以减少到：

t.right.left = lr

甚至：

n.left = lr

最终代码

通过上述更改，您的函数可以是：

class Node:
    def __init__(self, value):
        self.key = value
        self.left = None
        self.right = None
        self.parent = None

def rightRotate(node):
    if node is None or node.left is None:
        return node
    parent = node.parent
    left = node.left
    left_right = left.right

    # change edge 1
    if parent: # find out if node is a left or right child of node
        if parent.left == node:
            parent.left = left
        else:
            parent.right = left
    left.parent = parent

    # change edge 2
    left.right = node
    node.parent = left

    # change edge 3
    node.left = left_right
    if left_right:
        left_right.parent = node

    return left  # the node that took the position of node

# your code to build the tree
n_12 = Node(12)
n_15 = Node(15)
n_3 = Node(3)
n_7 = Node(7)
n_1 = Node(1)
n_2 = Node(2)
n_not1 = Node(-1)

n_12.right = n_15
n_12.left = n_3
n_3.right = n_7
n_3.left = n_1
n_1.right = n_2
n_1.left = n_not1

n_12.parent = None
n_15.parent = n_12
n_3.parent = n_12
n_7.parent = n_3
n_1.parent = n_3
n_2.parent = n_1
n_not1.parent = n_1

# rotate the root
root = n_12
root = rightRotate(root) # returns the node that took the place of n_12

只需删除带有 parent 的行即可获得 single-threaded 版本。

python 中树的左右旋转

Right and left rotation of a tree in python

python

tree

rotation

1) 赋值应该是无条件的

2) 双线程

其他备注：

最终代码