树的一个显著特点是对于节点 (x, y) 的每种组合,存在一条独特的非重复路径从 x 到 y。特别地,可以通过找到 x 和 y 的第一个共同祖先 z 并取路径 x->z+z->y 来找到此路径。
因此,查找所有路径的算法可能如下所示:
for each pair of distinct nodes x, y in the tree:
find all common ancestors of x and y
let z be the lowest common acestor in the tree
let p be the path from x to z
append the path from z to y to p, excluding the duplicate z
print p
这里提供了一种面向对象的方法,实现所需的方法以完成上述任务。
class Tree:
def __init__(self, value, left=None, right=None):
self.value = value
self.left = left
self.right = right
self.parent = None
if left is not None:
left.parent = self
if right is not None:
right.parent = self
def __iter__(self):
"""Return a left-to-right iterator on the tree"""
if self.left:
yield from iter(self.left)
yield self
if self.right:
yield from iter(self.right)
def __repr__(self):
return str(self.value)
def get_ancestors(self):
"""Return a list of all ancestors including itself"""
ancestors = {self}
parent = self.parent
while parent:
ancestors.add(parent)
parent = parent.parent
return ancestors
def get_path_to_ancestor(self, ancestor):
"""
Return the path from self to ancestor as a list
output format: [self, ..., ancestor]
"""
path = []
parent = self
try:
while True:
path.append(parent)
if parent is ancestor:
break
else:
parent = parent.parent
except AttributeError:
return None
return path
def get_path_to(self, other):
"""
Return the path from self to other as a list
output format: [self, ..., first common acestor, ..., other]
"""
common_ancestors = self.get_ancestors() & other.get_ancestors()
first_common_ancestor = {
a for a in common_ancestors
if a.left not in common_ancestors and a.right not in common_ancestors
}.pop()
return self.get_path_to_ancestor(first_common_ancestor)\
+ list(reversed(other.get_path_to_ancestor(first_common_ancestor)))[1:]
以下是针对您提供的树作为示例的应用方法。
tree = Tree(
6,
Tree(4,
Tree(1),
Tree(3)),
Tree(0,
Tree(1)))
nodes = list(tree)
for i in range(len(nodes)):
for j in range(i + 1, len(nodes)):
print([t for t in nodes[i].get_path_to(nodes[j])])
以下是所有被打印的路径。
[1, 4]
[1, 4, 3]
[1, 4, 6]
[1, 4, 6, 0, 1]
[1, 4, 6, 0]
[4, 3]
[4, 6]
[4, 6, 0, 1]
[4, 6, 0]
[3, 4, 6]
[3, 4, 6, 0, 1]
[3, 4, 6, 0]
[6, 0, 1]
[6, 0]
[1, 0]
4,6,0
是有效的吗?也就是说,你是要找到所有从根节点到叶子节点的路径以及所有叶子节点之间的路径,还是要找到任意两个节点之间的所有可能路径? - Zachary Cross6
个节点,您想要6*5//2 == 15
条路径。这正确吗?此外,请展示您用于二叉树的数据结构--这将影响任何答案。 - Rory Daulton