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二叉查找树:Python实现

2013-08-23 19:12 344 查看
#coding:utf8
#author:HaxtraZ

class BST(object):
"""二叉查找树的简单实现"""
def __init__(self):
self.root = None

def insert(self, val):
newNode = BSTnode(val)
if self.root is None:
self.root = newNode
else:
curNode = self.root
while True:
if val < curNode.val:
#进入左子树
if curNode.left is None:
curNode.left = newNode
newNode.parent = curNode
break
curNode = curNode.left
else:
#进入右子树
if curNode.right is None:
curNode.right = newNode
newNode.parent = curNode
break
curNode = curNode.right

def find(self, val):
curNode = self.root
while curNode is not None:
if val < curNode.val:
curNode = curNode.left
elif val > curNode.val:
curNode = curNode.right
else:
return True  # 找到了!

return False  # 没找到

def delete(self, val):
curNode = self.root
while curNode is not None:
if val < curNode.val:
curNode = curNode.left
elif val > curNode.val:
curNode = curNode.right
else:
# 找到了val
if curNode.left is not None and curNode.right is not None:
target = self.successor(curNode.right).val
curNode.val = target.val
self.delete(target)
elif curNode.left is not None:
if curNode == self.root:
self.root = curNode.left
parNode = curNode.parent
subNode = curNode.left
if parNode.left == curNode:
parNode.left = subNode
else:
parNode.right = subNode
subNode.parent = parNode
else:
if curNode == self.root:
self.root = curNode.right
parNode = curNode.parent
subNode = curNode.right
if parNode.right == curNode:
parNode.right = subNode
else:
parNode.left = subNode

return True
return False  # 不存在val,删除失败

def minimum(self, node):
# 返回最小值的节点。其实就是most left one
curNode = node
if curNode is None:
#空树
return None
while curNode.left is not None:
curNode = curNode.left
return curNode

def maximum(self, node):
#返回最大值的节点。其实就是most right one
curNode = node
if curNode is None:
#空树
return None
while curNode.right is not None:
curNode = curNode.right
return curNode

def successor(self, node):
#node是二叉查找树中的一个节点
#寻找node的后继节点,然后返回
curNode = node
if curNode.right is not None:
#右子树非空,返回右子树最左节点
return self.minimun(curNode.right)
else:
#右子树为空,则返回“最低祖先”
parNode = curNode.parent
while parNode is not None and parNode.right == curNode:
curNode = parNode
parNode = parNode.parent
return parNode

def show(self):
# 中序遍历
self.display(self.root)
print '\n'

def display(self, node):
if node is None:
return
self.display(node.left)
print node.val,
self.display(node.right)

#    def predecessor(self, node):
# 获取前驱节点。类似于获取后继节点,这里省略。。

class BSTnode(object):
"""二叉查找树的节点类型"""
def __init__(self, val):
self.val = val
self.left = None
self.right = None
self.parent = None

if __name__ == '__main__':
mylist = [2, 2, 7, 4, 1, 5]
bst = BST()
for i in range(len(mylist)):
bst.insert(mylist[i])

bst.show()
bst.delete(7)
bst.show()


二叉查找树,英文Binary Search Tree,也叫二叉排序树,是一种基本的数据结构,简称BST

它支持多种动态集合操作,包括查找(find),最小值(minimum),最大值(maximum),后继(successor),前驱(predecessor),插入(insert),删除(delete),以及中序遍历等。它既可以用作字典,也可以用作优先队列。

BST不是稳定的树,极端情况下会退化为线性结构,但平均期望上,insert,delete操作可以为O(lg n),树的期望高度为O(lg n)。

参考了《算法导论》等书,写出了具有insert,delete,find功能的BST,如果有认为不正确的地方欢迎拍砖。
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