Given 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.
Note: Time complexity should be O(height of tree).
Example:
root = [5,3,6,2,4,null,7]
key = 3
5
/ \
3 6
/ \ \
2 4 7
Given key to delete is 3. So we find the node with value 3 and delete it.
One valid answer is [5,4,6,2,null,null,7], shown in the following BST.
5
/ \
4 6
/ \
2 7
Another valid answer is [5,2,6,null,4,null,7].
5
/ \
2 6
\ \
4 7
"""
classSolution(object):
defdelete_node(self, root, key):
"""
:type root: TreeNode
:type key: int
:rtype: TreeNode
"""
ifnotroot: returnNone
ifroot.val==key:
ifroot.left:
# Find the right most leaf of the left sub-tree
left_right_most=root.left
whileleft_right_most.right:
left_right_most=left_right_most.right
# Attach right child to the right of that leaf
left_right_most.right=root.right
# Return left child instead of root, a.k.a delete root
returnroot.left
else:
returnroot.right
# If left or right child got deleted, the returned root is the child of the deleted node.