数据结构和算法-二叉查找树

二叉查找树(Binary Search Tree), 简称BST,必须具备如下性质:

• 若任意节点的左子树不空，则左子树上全部节点的值均小于它的根结点的值
• 若任意节点的右子树不空，则右子树上全部节点的值均大于它的根结点的值
• 任意节点的左、右子树也分别为二叉查找树
• 没有键值相等的节点

在二叉查找树中查找节点时, 平均运行时间为O(logn)(平衡状况), 最坏为O(n)(极度不平衡), 平均深度是O(logn)

在有序数组中使用二分查找时最坏的时间复杂度是O(logn), 可是二叉搜索树的插入和删除的性能更优

二叉搜索树和数组对比

-	数组	二叉搜索树
查找	O(logn)(二分查找)	O(logn)
插入	O(n)	O(logn)
删除	O(n)	O(logn)

实现

• 插入
• 查找
• 删除
  • 要删除的节点是叶子节点, 只须要让父节点指向null
  • 要删除的节点是只有一个子节点(无论是左/右), 须要让父节点指向子节点
  • 要删除的节点有两个子节点, 则须要找到该节点右子树中的最小节点, 而后替换到该节点, 而且再删除这个最小节点

特色

• 快速的插入/查找/删除
• 快速找到最大/最小节点
• 中序遍历能够获得有序的数据, O(n)

# coding:utf-8


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

    def isLeftChild(self):
        return self.parent and self.parent.left == self

    def isRightChild(self):
        return self.parent and self.parent.right == self

    def childrenNums(self):
        num = 0
        if self.left:
            num += 1
        if self.right:
            num += 1
        return num


class BinarySearchTree(object):
    def __init__(self):
        self.root = None
        self.size = 0

    def put(self, key, value):
        if self.root:
            self._put(key, value, self.root)
        else:
            self.root = TreeNode(key, value)
        self.size += 1

    def _put(self, key, value, current):
        if key < current.key:
            if current.left:
                self._put(key, value, current.left)
            else:
                current.left = TreeNode(key, value, parent=current)
        elif key > current.key:
            if current.right:
                self._put(key, value, current.right)
            else:
                current.right = TreeNode(key, value, parent=current)

    def get(self, key):
        if self.root:
            res = self._get(key, self.root)
        else:
            return None
        return res

    def _get(self, key, current):
        if not current:
            return None
        if key < current.key:
            return self._get(key, current.left)
        elif key > current.key:
            return self._get(key, current.right)
        else:
            return current

    def delete(self, key):
        if self.size > 1:
            # 要先找到该节点
            node2remove = self.get(key)
            if node2remove:
                self.remove(node2remove)
                self.size -= 1
            else:
                raise Exception('no element')
        elif self.size == 1 and self.root.key == key:
            self.root = None
            self.size -= 1
        else:
            raise Exception('no element')

    def remove(self, current):
        childrens = current.childrenNums()
        if 0 == childrens:
            if current.isLeftChild():
                current.parent.left = None
            else:
                current.parent.right = None
        elif 1 == childrens:
            # 若是该节点有左子树
            if current.left:
                if current.isLeftChild():
                    current.left.parent = current.parent
                    current.parent.left = current.left
                elif current.isRightChild():
                    current.left.parent = current.parent
                    current.parent.right = current.left
                else:
                    self.root = current.left
            # 若是是右子树
            elif current.right:
                if current.isLeftChild():
                    current.right.parent = current.parent
                    current.parent.left = current.right
                elif current.isRightChild():
                    current.right.parent = current.parent
                    current.parent.right = current.right
                else:
                    self.root = current.right
        # 若是有两个子节点
        else:
            parent = current
            minChild = current.right
            while minChild.left != None:
                parent = minChild
                minChild = minChild.left
            current.key = minChild.key
            current.value = minChild.value
            # 注意如下状况判断, 由于有的没有左子节点
            if parent.left == minChild:
                parent.left = minChild.right
                if minChild.right:
                    minChild.right.parent = parent
            else:
                parent.right = minChild.right
                if minChild.right:
                    minChild.right.parent = parent

    def length(self):
        return self.size

    def __setitem__(self, key, value):
        self.put(key, value)

    def __getitem__(self, key):
        return self.get(key)

    def __delitem__(self, key):
        self.delete(key)

    def mid(self, root):
        if not root:
            return
        self.mid(root.left)
        print(root.value)
        self.mid(root.right)



if __name__ == '__main__':
    mytree = BinarySearchTree()
    mytree[7] = "7"
    mytree[2] = "2"
    mytree[11] = "11"
    mytree[8] = "8"
    mytree[19] = "19"
    mytree[5] = "5"
    mytree[1] = "1"

    # 中序遍历
    mytree.mid(mytree.root)
    print('------')

    # 删除叶子节点1
    del mytree[1]
    mytree.mid(mytree.root)
    print('------')

    # 删除有一个子节点的2
    del mytree[2]
    mytree.mid(mytree.root)
    print('------')

    # 删除有两个子节点的11
    del mytree[11]
    mytree.mid(mytree.root)

时间复杂度

• 最坏: 退化成链表 O(n)
• 最好: 平衡 O(logn)

问题

散列表更加高效, 为何还有二叉查找树?

• 散列表的数据时无序存储的, 若是要输出有序数据的话还须要额外进行排序. 可是二叉查找树能够直接中序遍历(O(n))
• 散列表扩容耗时长, 遇到散列冲突时性能不稳定. 工程中用的AVL树性能稳定
• 散列表构造复杂, 须要考虑散列函数, 冲突解决等. 而二叉查找树仅仅考虑平衡问题

资料

• < <大话数据结构> >
• 二叉树
• < <数据结构和算法> >