tasks for today:
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699.修建二叉搜索树
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108.将有序数组转化为二叉搜索树
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538.把二叉搜索树转化为累加树
- 699.修建二叉搜索树
IN this practice, the feature of being binary search tree is very important, this can help trim the tree speedily.
when in "if root.val < low:" this condition, the left child tree can be totally trimed, because this is a binary search tree. All the left child value is less than the value of current root.
Note: Please compare the difference between this practice and the practice 450.
python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def trimBST(self, root: Optional[TreeNode], low: int, high: int) -> Optional[TreeNode]:
if root is None:
return None
if root.val < low:
# when in this condition, the left child tree can be totally trimed, because this is a binary search tree
return self.trimBST(root.right, low, high)
elif root.val > high:
# when in this condition, the right child tree can be totally trimed, because this is a binary search tree
return self.trimBST(root.left, low, high)
root.left = self.trimBST(root.left, low, high)
root.right = self.trimBST(root.right, low, high)
return rootone question: if I follow the logic of practice 450, it sometimes work well, but sometimes return fause solution, I mean: it is true that if a node's value is outside the range [low, high], the entire subtree rooted at that node needs to be trimmed, but I wonder, if I follow the logic of deleting a node to trim the tree, although it is more clumsy, but that should also work, the difference is I transaction the trimming a subtree into trimming nodes one-buy-one, why it sometimes not work, and sometimes work.
This problem might be created by the traverse order issue, because based on the example caser, some removed node is not correctly remove becasue some pitential trace back in the recursive algorithm. [2,1,3], low=3, high=4, [3] is expected but [3,1] is returned.
But this problem is not in delete node, maybe because the judging condition of root.val > key or root.val < key.
- 108.将有序数组转化为二叉搜索树
this practice's target is build a tree, so the key idea is to identify the val for construct current root node and the corresponding list feed into it for its following branches construction.
python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
if not nums:
return
new_node = TreeNode(nums[int((len(nums)-1)/2)])
new_node.left = self.sortedArrayToBST(nums[:int((len(nums)-1)/2)])
new_node.right = self.sortedArrayToBST(nums[int((len(nums)-1)/2)+1:])
return new_node- 538.把二叉搜索树转化为累加树
for a binary seach tree, the inorder search is a ascending list, to calculate the sum of values larger than cur node, the inorder shoudl be reversed, which makes a descending list, by adding the nums before the value of current node, the cur tree node's value can be given to cur root node for upate.
Noted: when there are variables in recursive, a self.XXX definition should be necessary.
python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
# when there is variable in the recursive, this from should be used
self.pre = 0
self.rev_inorder(root)
return root
def rev_inorder(self, node):
if node is None:
return
self.rev_inorder(node.right)
node.val += self.pre
self.pre = node.val
self.rev_inorder(node.left)