binary search tree

它写的很好一点，就是大小写分的很清，我经常都是一边写。这个位置就要学习下。代码抽空看看。官网的二分源码。还是值得学习的。多学习下优秀的代码。

 　　　  if key < node.key:
            return self._get(key, node.left)
        elif key > node.key:
            return self._get(key, node.right)
        else:
            return node.val

而且都是在前面定义 了属性，再定义方法。把这两者分开。我就没有，我不确定，因为前面都是_get , 后面才又跟着一个get。

构造函数用单一的前置下划线（_）来创建属性。属性名以单下划线开头。虽然在python语法中没有特殊的含义，但单下划线是python程序员使用类是约定的使用的符号，表明程序员不希望类的用户直接访问属性。如果程序要求访问属性，程序员会提供其他途径

"""
    Binary Search Tree
    ------------------
    The Binary Search Tree represents an ordered symbol table of generic
    key-value pairs.  Keys must be comparable.  Does not permit duplicate keys.
    When assocating a value with a key already present in the BST, the previous
    value is replaced by the new one.  This implementation is for an unbalanced
    BST.

    Pseudo Code: http://algs4.cs.princeton.edu/32bst
"""


[docs]class Node(object):
    """
    Implementation of a Node in a Binary Search Tree.
    """

    def __init__(self, key=None, val=None, size_of_subtree=1):
        self.key = key
        self.val = val
        self.size_of_subtree = size_of_subtree
        self.left = None
        self.right = None


[docs]class BinarySearchTree(object):
    """
    Implementation of a Binary Search Tree.
    """

    def __init__(self):
        self.root = None

    def _size(self, node):
        if node is None:
            return 0
        else:
            return node.size_of_subtree

[docs]    def size(self):
        """
        Return the number of nodes in the BST

        Worst Case Complexity: O(1)

        Balanced Tree Complexity: O(1)
        """
        return self._size(self.root)

[docs]    def is_empty(self):
        """
        Returns True if the BST is empty, False otherwise

        Worst Case Complexity: O(1)

        Balanced Tree Complexity: O(1)
        """
        return self.size() == 0

    def _get(self, key, node):
        if node is None:
            return None

        if key < node.key:
            return self._get(key, node.left)
        elif key > node.key:
            return self._get(key, node.right)
        else:
            return node.val

[docs]    def get(self, key):
        """
        Return the value paired with 'key'

        Worst Case Complexity: O(N)

        Balanced Tree Complexity: O(lg N)
        """
        return self._get(key, self.root)

[docs]    def contains(self, key):
        """
        Returns True if the BST contains 'key', False otherwise

        Worst Case Complexity: O(N)

        Balanced Tree Complexity: O(lg N)
        """
        return self.get(key) is not None

    def _put(self, key, val, node):

        # If we hit the end of a branch, create a new node
        if node is None:
            return Node(key, val)

        # Follow left branch
        if key < node.key:
            node.left = self._put(key, val, node.left)
        # Follow right branch
        elif key > node.key:
            node.right = self._put(key, val, node.right)
        # Overwrite value
        else:
            node.val = val

        node.size_of_subtree = self._size(node.left) + self._size(node.right)+1
        return node

[docs]    def put(self, key, val):
        """
        Add a new key-value pair.

        Worst Case Complexity: O(N)

        Balanced Tree Complexity: O(lg N)
        """
        self.root = self._put(key, val, self.root)

    def _min_node(self):
        """
        Return the node with the minimum key in the BST
        """
        min_node = self.root
        # Return none if empty BST
        if min_node is None:
            return None

        while min_node.left is not None:
            min_node = min_node.left

        return min_node

[docs]    def min_key(self):
        """
        Return the minimum key in the BST

        Worst Case Complexity: O(N)

        Balanced Tree Complexity: O(lg N)
        """
        min_node = self._min_node()
        if min_node is None:
            return None
        else:
            return min_node.key

    def _max_node(self):
        """
        Return the node with the maximum key in the BST
        """
        max_node = self.root
        # Return none if empty BST
        if max_node is None:
            return None

        while max_node.right is not None:
            max_node = max_node.right

        return max_node

[docs]    def max_key(self):
        """
        Return the maximum key in the BST

        Worst Case Complexity: O(N)

        Balanced Tree Complexity: O(lg N)
        """
        max_node = self._max_node()
        if max_node is None:
            return None
        else:
            return max_node.key

    def _floor_node(self, key, node):
        """
        Returns the node with the biggest key that is less than or equal to the
        given value 'key'
        """
        if node is None:
            return None

        if key < node.key:
            # Floor must be in left subtree
            return self._floor_node(key, node.left)

        elif key > node.key:
            # Floor is either in right subtree or is this node
            attempt_in_right = self._floor_node(key, node.right)
            if attempt_in_right is None:
                return node
            else:
                return attempt_in_right

        else:
            # Keys are equal so floor is node with this key
            return node

[docs]    def floor_key(self, key):
        """
        Returns the biggest key that is less than or equal to the given value
        'key'

        Worst Case Complexity: O(N)

        Balanced Tree Complexity: O(lg N)
        """
        floor_node = self._floor_node(key, self.root)
        if floor_node is None:
            return None
        else:
            return floor_node.key

    def _ceiling_node(self, key, node):
        """
        Returns the node with the smallest key that is greater than or equal to
        the given value 'key'
        """
        if node is None:
            return None

        if key < node.key:
            # Ceiling is either in left subtree or is this node
            attempt_in_left = self._ceiling_node(key, node.left)
            if attempt_in_left is None:
                return node
            else:
                return attempt_in_left
        elif key > node.key:
            # Ceiling must be in right subtree
            return self._ceiling_node(key, node.right)
        else:
            # Keys are equal so ceiling is node with this key
            return node

[docs]    def ceiling_key(self, key):
        """
        Returns the smallest key that is greater than or equal to the given
        value 'key'

        Worst Case Complexity: O(N)

        Balanced Tree Complexity: O(lg N)
        """
        ceiling_node = self._ceiling_node(key, self.root)
        if ceiling_node is None:
            return None
        else:
            return ceiling_node.key

    def _select_node(self, rank, node):
        """
        Return the node with rank equal to 'rank'
        """
        if node is None:
            return None

        left_size = self._size(node.left)
        if left_size < rank:
            return self._select_node(rank - left_size - 1, node.right)
        elif left_size > rank:
            return self._select_node(rank, node.left)
        else:
            return node

[docs]    def select_key(self, rank):
        """
        Return the key with rank equal to 'rank'

        Worst Case Complexity: O(N)

        Balanced Tree Complexity: O(lg N)
        """
        select_node = self._select_node(rank, self.root)
        if select_node is None:
            return None
        else:
            return select_node.key

    def _rank(self, key, node):
        if node is None:
            return None

        if key < node.key:
            return self._rank(key, node.left)
        elif key > node.key:
            return self._size(node.left) + self._rank(key, node.right) + 1

        else:
            return self._size(node.left)

[docs]    def rank(self, key):
        """
        Return the number of keys less than a given 'key'.

        Worst Case Complexity: O(N)

        Balanced Tree Complexity: O(lg N)
        """
        return self._rank(key, self.root)

    def _delete(self, key, node):
        if node is None:
            return None
        if key < node.key:
            node.left = self._delete(key, node.left)
        elif key > node.key:
            node.right = self._delete(key, node.right)

        else:
            if node.right is None:
                return node.left
            elif node.left is None:
                return node.right
            else:
                old_node = node
                node = self._ceiling_node(key, node.right)
                node.right = self._delete_min(old_node.right)
                node.left = old_node.left
        node.size_of_subtree = self._size(node.left) + self._size(node.right)+1
        return node

[docs]    def delete(self, key):
        """
        Remove the node with key equal to 'key'

        Worst Case Complexity: O(N)

        Balanced Tree Complexity: O(lg N)
        """
        self.root = self._delete(key, self.root)

    def _delete_min(self, node):
        if node.left is None:
            return node.right

        node.left = self._delete_min(node.left)
        node.size_of_subtree = self._size(node.left) + self._size(node.right)+1
        return node

[docs]    def delete_min(self):
        """
        Remove the key-value pair with the smallest key.


        Worst Case Complexity: O(N)

        Balanced Tree Complexity: O(lg N)
        """
        self.root = self._delete_min(self.root)

    def _delete_max(self, node):
        if node.right is None:
            return node.left

        node.right = self._delete_max(node.right)
        node.size_of_subtree = self._size(node.left) + self._size(node.right)+1
        return node

[docs]    def delete_max(self):
        """
        Remove the key-value pair with the largest key.

        Worst Case Complexity: O(N)

        Balanced Tree Complexity: O(lg N)
        """
        self.root = self._delete_max(self.root)

    def _keys(self, node, keys):
        if node is None:
            return keys

        if node.left is not None:
            keys = self._keys(node.left, keys)

        keys.append(node.key)

        if node.right is not None:
            keys = self._keys(node.right, keys)

        return keys

[docs]    def keys(self):
        """
        Return all of the keys in the BST in aschending order

        Worst Case Complexity: O(N)

        Balanced Tree Complexity: O(N)
        """
        keys = []
        return self._keys(self.root, keys)