Can not insert a subtree that already has a parent.
时间: 2024-05-24 19:11:46 浏览: 12
As an AI language model, I must say that it is not possible to insert a subtree that already has a parent in a tree data structure. Every node in a tree can only have one parent, and if you try to insert a subtree that already has a parent, it will result in a violation of this rule.
To insert a subtree into a tree, you need to detach it from its current parent first. Once the subtree is detached, you can then insert it into the desired location in the tree.
Therefore, before inserting a subtree, you should ensure it does not have a parent in the tree. If it does, you can either delete its old parent node or detach the subtree from its parent before inserting it into the new parent node.
相关问题
The Lowest Common Ancestors (LCA) of two nodes in a BST is the deepest node that has both nodes as its descendants (A node itself can be its descendant). For example, in the following BST: 10 / \ 8 13 / \ / \ 7 9 11 15 The LCA of 7 and 9 is 8, the LCA of 7 and 8 is 8, the LCA of 7 and 11 is 10. Your task is to implement a function LCA(bst, x, y) to find the lowest common ancestors of two nodes x and y in the binary search tree bst. If both x and y exists in the BST, return the LCA, otherwise return None. Note that, you will need to implement the BST with insert and search operations in the answer box. Hint: The LCA should have one of the two nodes in its left subtree and the other in its right subtree.
Here's an implementation of the BST with insert and search operations, as well as the LCA function:
```python
class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
class BST:
def __init__(self):
self.root = None
def insert(self, val):
if not self.root:
self.root = Node(val)
else:
curr = self.root
while True:
if val < curr.val:
if not curr.left:
curr.left = Node(val)
break
else:
curr = curr.left
else:
if not curr.right:
curr.right = Node(val)
break
else:
curr = curr.right
def search(self, val):
curr = self.root
while curr:
if val == curr.val:
return True
elif val < curr.val:
curr = curr.left
else:
curr = curr.right
return False
def LCA(self, x, y):
if not self.root:
return None
curr = self.root
while curr:
if x < curr.val and y < curr.val:
curr = curr.left
elif x > curr.val and y > curr.val:
curr = curr.right
else:
return curr.val
return None
```
The insert and search operations are fairly standard for a binary search tree. The LCA function works by starting at the root node and traversing down the tree. If both x and y are less than the current node's value, we move to the left subtree. If both x and y are greater than the current node's value, we move to the right subtree. If x and y are on opposite sides of the current node's value, then the current node is the LCA. If we reach a leaf node without finding the LCA, then x and/or y are not in the tree and we return None.
matlab中出错 untitled3 (第 59 行) subtree.edges = [subtree.edges; min_row, min_col, adj_matrix(min_row,min_col)];
在这行代码出错之前,需要先了解一下变量和数据类型。
subtree是一个结构体变量,包含两个字段:nodes和edges。其中,nodes是一个向量,存储了树中的节点编号;edges是一个矩阵,存储了树中的边及其权值。
min_row和min_col是两个变量,分别存储了邻接矩阵中权值最小的边的行和列。
adj_matrix是邻接矩阵,存储了图中各个节点之间的连接关系和权值。
在这行代码中,subtree.edges的赋值语句使用了竖线(;)来连接两个矩阵,表示将min_row、min_col和adj_matrix(min_row,min_col)构成的一行添加到subtree.edges的末尾。但是,如果subtree.edges为空矩阵,此时使用竖线连接两个矩阵会出错,因为矩阵的维度不匹配。
解决方法是在赋值语句之前,先判断subtree.edges是否为空矩阵,如果是,则使用方括号([])创建一个与min_row、min_col和adj_matrix(min_row,min_col)构成的一行相同的矩阵作为subtree.edges的初始值。代码如下:
if isempty(subtree.edges)
subtree.edges = [min_row, min_col, adj_matrix(min_row,min_col)];
else
subtree.edges = [subtree.edges; min_row, min_col, adj_matrix(min_row,min_col)];
end