Delete Node in BST

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

Solution

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def deleteNode(self, root, key):
        """
        :type root: TreeNode
        :type key: int
        :rtype: TreeNode
        """
        if root is None: return
        if root.val == key:
            if root.right and root.left:
                right = root.right
                while right is not None and right.left is not None: right = right.left
                right.left = root.left
                return root.right
            else:
                if root.left:
                    return root.left
                elif root.right:
                    return root.right
                else:
                    return None
        elif key > root.val:
            root.right = self.deleteNode(root.right, key)
        elif key < root.val:
            root.left = self.deleteNode(root.left, key)

        return root