Medium
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.
Example 1:
Input: root = [5,3,6,2,4,null,7], key = 3 Output: [5,4,6,2,null,null,7] Explanation: 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 above BST. Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted.
Example 2:
Input: root = [5,3,6,2,4,null,7], key = 0 Output: [5,3,6,2,4,null,7] Explanation: The tree does not contain a node with value = 0.
Example 3:
Input: root = [], key = 0 Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 104]. -105 <= Node.val <= 105- Each node has a unique value.
rootis a valid binary search tree.-105 <= key <= 105
Follow up: Could you solve it with time complexity O(height of tree)?
# 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 successor(self, root):
""" Successor is the smallest child of the right node. """
root = root.right
while root.left:
root = root.left
return root.val
def predecessor(self, root):
""" Predecessor is the largest child of the left node. """
root = root.left
while root.right:
root = root.right
return root.val
def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
if not root:
return None
# let's find the node, is it in left or right or just current node
if key > root.val:
root.right = self.deleteNode(root.right, key)
elif key < root.val:
root.left = self.deleteNode(root.left, key)
else: # found it!!!!
# is it a leaf node?
if not root.left and not root.right:
root = None # deleted!!!
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