Question

是否有这样的数字序列（1-7，所有使用的数字，每次只有一次），会形成相等的AVL和splay树？

Answer 1

嗯，为了科学的利益，我根据各自的维基百科文章在Python中实现了AVL和splay树。假设我没有在某处犯错，我的发现是没有{1，...，7}的排列产生相同的AVL和splay树。我猜想对于所有尺寸为k的组都是如此。至于这个的根本原因，我不知道。

如果有人想审核我的代码，请点击：

#####################
# Class definitions #
#####################
class Node:
    ''' A binary tree node '''
    def __init__(self, n, p=None, l=None, r=None):
        self.n = n
        self.p = p
        self.l = l
        self.r = r
        self.h = None

    def __str__(self):
        return "[%s %s %s]" % (self.n, (self.l if self.l else "-"), (self.r if self.r else "-"))

    def __eq__(self, other):
        if not isinstance(other, Node):
            return NotImplemented
        return self.n == other.n and self.l == other.l and self.r == other.r

    def __ne__(self, other):
        return not (self == other)

class HNode(Node):
    ''' A binary tree node, with height '''
    def __init__(self, n, p=None, l=None, r=None):
        super(HNode, self).__init__(n, p, l, r)
        self.hup()

    def lh(self):
        ''' Get height of left child '''
        return self.l.h if self.l else 0

    def rh(self):
        ''' Get height of right child '''
        return self.r.h if self.r else 0

    def hup(self):
        ''' Update node height '''
        self.h = max(self.lh(), self.rh())+1

    def __str__(self):
        return "[%s (%d) %s %s]" % (self.n, self.h, (self.l if self.l else "-"), (self.r if self.r else "-"))

#########################
# Basic tree operations #
#########################

#      v           u
#     / \         / \
#    u   c  -->  a   v
#   / \             / \
#  a   b           b   c
def rright(v):
    ''' Rotate right '''
    u = v.l
    u.r, v.l = v, u.r
    if v.l:
        v.l.p = v
    u.p, v.p = v.p, u
    return u

#    u             v
#   / \           / \
#  a   v   -->   u   c
#     / \       / \
#    b   c     a   b
def rleft(u):
    ''' Rotate left '''
    v = u.r
    u.r, v.l = v.l, u
    if u.r:
        u.r.p = u
    u.p, v.p = v, u.p
    return v

######################
# AVL tree functions #
######################

def avl_lr(v):
    v.l = rleft(v.l)
    v.l.l.hup()
    v.l.hup()
    return avl_ll(v)

def avl_ll(v):
    u = rright(v)
    u.r.hup()
    u.hup()
    return u

def avl_rl(v):
    v.r = rright(v.r)
    v.r.r.hup()
    v.r.hup()
    return avl_rr(v)

def avl_rr(v):
    u = rleft(v)
    u.l.hup()
    u.hup()
    return u

def avl_insert(v, n, p=None):
    if v is None:
        return HNode(n, p)
    if n < v.n:
        v.l = avl_insert(v.l, n, v)
        v.hup()
        if v.lh() > v.rh() + 1:
            return (avl_ll if (v.l.lh() > v.l.rh()) else avl_lr)(v)
        else:
            return v
    else:
        v.r = avl_insert(v.r, n, v)
        v.hup()
        if v.rh() > v.lh() + 1:
            return (avl_rr if (v.r.rh() > v.r.lh()) else avl_rl)(v)
        else:
            return v

def build_avl_tree(s):
    ''' Build an AVL tree from the given sequence '''
    v = None
    for n in s:
        v = avl_insert(v, n)
    return v

########################
# Splay tree functions #
########################

two = lambda x: (x,x)

def bst_insert(p, n, g=None):
    ''' Insert a value into a BST, returning a pair consisting of
        the root of the tree and the new node '''
    if p is None:
        return two(Node(n,g))
    if n < p.n:
        p.l, x = bst_insert(p.l, n, p)
    else:
        p.r, x = bst_insert(p.r, n, p)
    return p, x

def splay(x):
    ''' Percolate x to the root of its tree '''
    if x.p:
        p = x.p
        g = p.p
        if g:
            if p.n < g.n:
                if x.n < p.n:
                    x = rright(rright(g))
                else:
                    g.l = rleft(p)
                    x = rright(g)
            else:
                if x.n > p.n:
                    x = rleft(rleft(g))
                else:
                    g.r = rright(p)
                    x = rleft(g)
            p = x.p
            if p:
                if x.n < p.n:
                    p.l = x
                else:
                    p.r = x
            return splay(x)
        else:
            if x.n < p.n:
                return rright(p)
            else:
                return rleft(p)
    else:
        return x

def splay_insert(p, n, g=None):
    r, x = bst_insert(p, n, g)
    return splay(x)

def build_splay_tree(s):
    ''' Build a splay tree from the given sequence '''
    v = None
    for n in s:
        v = splay_insert(v, n)
    return v

####################
# The Big Question #
####################

from itertools import permutations
def find_matches(n):
    ''' Generate all permutations of {1, ..., n} that produce
        matching AVL and splay trees '''
    for s in permutations(range(1, n+1)):
        t1 = build_avl_tree(s)
        t2 = build_splay_tree(s)
        if t1 == t2:
            yield s

def find_match(n):
    ''' Return a permutation of {1, ..., n} that produces matching
        AVL and splay trees, or None if no such permutation exists '''
    return next(find_matches(n), None)

一个形成相同AVL和splay树的序列？

1 个答案: