450. Delete Node in a BST
Category: /leetcodeGiven a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.
Basically, the deletion can be divided into two stages:
Search for a node to remove. If the node is found, delete the node. Follow up: Can you solve it with time complexity O(height of tree)?
Solution:
def successor(self, root):
root = root.right
while root.left:
root = root.left
return root.val
def predecessor(self, root):
root = root.left
while root.right:
root = root.right
return root.val
def deleteNode(self, root: TreeNode, key: int) -> TreeNode:
if not root:
return None
# delete from the right subtree
if key > root.val:
root.right = self.deleteNode(root.right, key)
elif key < root.val:
root.left = self.deleteNode(root.left, key)
else: # found the node
if not (root.left or root.right):
# the node is a leaf
root = None
elif root.right:
root.val = self.successor(root)
root.right = self.deleteNode(root.right, root.val)
else:
root.val = self.predecessor(root)
root.left = self.deleteNode(root.left, root.val)
return root
