450. Delete Node In A Bst

450. Delete Node in a BST

Medium

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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:

1. Search for a node to remove.
2. 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.
• root is 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