二叉树查找

"""
二叉查找树：BST：
    每个节点包含一个`key`和他附带的数据，对于每个内部节点`V`
    所有`key`都小于`V`的都被存储在`V`的左子树
    所有`key`小于`V`的都存储在`V`的右子树
    
时间复杂度：O(n)
"""

class BSTNode(object):
    def __init__(self, key, value, left=None, right=None):
        """
            left指向左子树
            right指向右子树
            None 表示没有左子树或右子树
        """
        self.key, self.value, self.left, self.right = key, value, left, right
        

class BST(object):
    def __init__(self, root=None):
        self.root = root
        
    def _bst_search(self, subtree, key):
        if subtree is None:
            return None
        elif key < subtree.key:
            return self._bst_search(subtree.left, key)
        elif key > subtree.key:
            return self._bst_search(subtree.right, key)
        else:
            return subtree
            
    def get(self, key, default=None):
        node = self._bst_search(self.root, key)
        if node is None:
            return default
        else:
            return node.value
            
    def __contains__(self, key):
        """实现 in 操作符"""
        return self._bst_search(self.root, key) is not None
        
    def _bst_min_node(self, subtree):
        if subtree is None:
            return None
        elif subtree.left is None:
            return subtree
        else:
            return self._bst_min_node(subtree.left)
            
    def bst_min(self):
        node = self._bst_min_node(self.root)
        return node.value if node else None
        
    def _bst_insert(self, subtree, key, value):
        if subtree is None:
            subtree = BSTNode(key, value)
        elif key < subtree.key:
            subtree.left = self._bst_insert(subtree.left, key, value)
        elif key > subtree.key:
            subtree.right = self._bst_insert(subtree.right, key, value)
        return subtree
        
    def add(self, key, value):
        node = self._bst_search(self.root, key)
        if node is not None:
            node.value = value
            return False
        else:
            self.root = self._bst_insert(self.root, key, value)
            self.size += 1
            return True
        
    @classmethod
    def build_from(cls, node_list):
        cls.size = 0
        key_to_node_dict = {}
        for node_dict in node_list:
            key = node_dict['key']
            key_to_node_dict[key] = BSTNode(key, value=key)  # 这里用key当作value
            
        for node_dict in node_list:
            key = node_dict['key']
            node = key_to_node_dict[key]
            if node_dict['is_root']:
                root = node
            node.left = key_to_node_dict.get(node_dict['left'])
            node.right = key_to_node_dict.get(node_dict['right'])
            cls.size += 1
        return cls(root)
        
    def _bst_remove(self, subtree, key):
        if subtree is None:
            return None
        elif key < subtree.key:
            subtree.left = self._bst_remove(subtree.left, key)
            return subtree
        elif key > subtree.key:
            subtree.right = self._bst_remove(subtree.right, key)
            return subtree
        else:  # 找到了需要删除的节点
            if subtree.left is None and subtree.right is None:
                # 叶子节点，返回 None，把其父亲指向它的指针
                return None
            elif subtree.left is None or subtree.right is None:
                # 只有一个孩子
                if subtree.left is not None:
                    return subtree.left
                else:
                    return subtree.right
            else:  # 两个孩子，需要寻找后继节点替换
                successor_node = self._bst_min_node(subtree.right)
                subtree.key, subtree.value = successor_node.key, successor_node.value
                subtree.right = self._bst_remove(subtree.right, successor_node.key)
                return subtree
                
    def remove(self, key):
        """
            中(根)序遍历二叉查找树，会得到一个有序的数组，从而，得到了两个概念，逻辑前任，逻辑后继
            逻辑前任：有序数组中此节点的上一个节点
            逻辑后继：下一个节点
            并且，产生一个规律：逻辑后继，总是此节点右子树的最小值，原因：二叉树的性质，右边的树都比此节点大，所以右边的
        值肯定都大于此节点，所以左子树的最小值就是后继
        """
        assert key in self
        self.size -= 1
        return self._bst_remove(self.root, key)
        
NODE_LIST = [
    {'key': 60, 'left': 12, 'right': 90, 'is_root':True},
    {'key': 12, 'left': 4, 'right': 41, 'is_root':False},
    {'key': 4, 'left': 1, 'right': None, 'is_root':False},
    {'key': 1, 'left': None, 'right': None, 'is_root':False},
    {'key': 41, 'left': 29, 'right': None, 'is_root':False},
    {'key': 29, 'left': 23, 'right': 37, 'is_root':False},
    {'key': 23, 'left': None, 'right': None, 'is_root':False},
    {'key': 37, 'left': None, 'right': None, 'is_root':False},
    {'key': 90, 'left': 71, 'right': 100, 'is_root':False},
    {'key': 71, 'left': None, 'right': 84, 'is_root':False},
    {'key': 100, 'left': None, 'right': None, 'is_root':False},
    {'key': 84, 'left': None, 'right': None, 'is_root':False},
]


def test_bst_tree():
    bst = BST.build_from(NODE_LIST)
    for node_dict in  NODE_LIST:
        key = node_dict['key']
        assert bst.get(key) == key
        
    assert bst.size == len(NODE_LIST)
    assert bst.get(-4) is None
    assert bst.bst_min() == 1
    
    bst.add(0, 0)
    assert bst.bst_min() == 0
    
    bst.remove(12)
    assert bst.get(12) is None
    
    bst.remove(1)
    assert bst.get(1) is None
    
    bst.remove(41)
    assert bst.get(41) is None
    
    bst.remove(29)
    assert bst.get(29) is None