python 数据结构之二叉搜索树

二叉搜索树定义

一颗二叉搜索树是以二叉树来组织的,每个节点除了 Key 还包括 左孩子, 右孩子, 父节点 等信息. BST满足限制条件: 对于任意节点的X,他的 左子树中关键字最大值<=X.key , 右子树关键字最小值>=X.key 这个关系表示如下

根据上图定义,一个二叉搜索树的例子是

二叉树操作

• 查询
• 插入
• 删除

查询(搜索)

二叉树搜索采用递归的方式来进行查询,根据二叉搜索树的定义: 左子树存储小值, 右子树存储大值,一个完整的二叉搜索示意图如下

可以写成 伪代码

TREE-SEARCH(x, k)  if x == NULL  or k == x.key    return x    if k < x.key    return TREE-SEARCH(x.left)  if k > x.key    return TREE-SEARCH(x.right)

转换成python代码

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.valdef 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)

插入

插入和删除比查询呢稍微复杂一些,因为该操作会引起二叉搜索树的大小变化,会改变动态集合的结构.插入呢又比删除稍微容易实现.插入分为两部

• 查询插入节点
• 改变目标节点附近的数据结构

插入过程示意图如下

相应的伪代码如下, 输入节点 z , z.key = v, z.left = NULL, z.right = NULL.

TREE-INSERT(T, x)  y = NULL  x = T.root   # 从根节点开始  while x != NULL    y = x      # 保存上一节点    if z.key < x.key # 往左      x = x.left    else             # 往右      x = x.right  z.p = y        # 父节点  if y == NULL   # tree T 为空    T.root = z  else if z.key < y.key    y.left = z  else y.right = z

程序的运行复杂度取决于二叉树的形状

插入的运行时间取决于二叉搜索树的高度h,程序的运行时间O(h) ,所以二叉树形状的好坏直接影响算法的运行时间.

python代码实现为

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 nodedef 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)

删除

删除总共分为三种情况:

• 如果删除节点x没有孩子,直接删除即可;
• 如果删除节点x有1个孩子,用孩子替换该节点位置;

• 如果删除节点x有2个孩子, 这个情况有些复杂.关键是要找到节点 x的继承者 . 节点z的继承者在节点z的右子树中有最小的关键值.这种情况下的操作分为下面步骤:

  1. 输入待删除的节点x 和 二叉搜索树T.
  2. 在节点x的右子树开始搜索:往右再往左找到最小值节点H;
  3. H右孩子为H的父节点, H的左孩子为X的左孩子;

示意图如下,应该一目了然:

根据上面的描述,删除的伪代码可以分为两部分:

1. 为了移动子树, 用一棵子树替换一棵子树,并成为双亲的孩子节点.

   TRANSPLANT(T, u, v)if u.p == NULLT.root = velse if u = u.p.leftu.p.left = velse u.p.right = vif v!= NULLv.p = u.p

2. 根据第一步完成二叉搜索树的删除过程:

   TREE-DELETE(T, z)if z.left = NULLTRANSPLANT(T, z, z.right)else if (z.right == NULL)TRANSPLANT(T, z, z.left)elsey = TREE-MINIMUM(z.right)if y.p != zTRANSPLANT(T, y, y.right)y.right = z.righty.right.p = yTRANSPLANT(T, z, y)y.left = z.lefty.left.p = y

   用python 实现如下:

   def _delete(self, key, node):if node is None:   return Noneif 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.leftnode.size_of_subtree = self._size(node.left) + self._size(node.right)+1return nodedef _delete_min(self, node):if node.left is None:   return node.rightnode.left = self._delete_min(node.left)node.size_of_subtree = self._size(node.left) + self._size(node.right)+1return nodedef _ceiling_node(self, key, node):"""Returns the node with the smallest key that is greater than or equal tothe given value 'key'"""if node is None:   return Noneif 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_leftelif 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

参考文献

1. <<算法导论第三版>>
2. http://algs4.cs.princeton.edu/32bst/