450. Delete Node in a 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:

1. Search for a node to remove.
2. 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

Approach 1: iterative.

Approach 2: recursive.

Code (Python)

Approach 1:

    def deleteNode(self, root, key):
        # iterative
        node = root
        parent = TreeNode(0)
        parent.right = root
        anchor = parent
        # search for the node
        while node:
            if node.val == key:
                break
            elif node.val < key:
                parent = node
                node = node.right
            else:
                parent = node
                node = node.left
        # node not found
        if not node:
            return root
        # node is a leaf -- just delete the node
        if not node.left and not node.right:
            if parent.left == node:
                parent.left = None
            else:
                parent.right = None
        # node has one child -- move its subtree up
        elif not node.left or not node.right:
            if parent.left == node:
                parent.left = node.left if node.left else node.right
            else:
                parent.right = node.left if node.left else node.right
        # node has 2 children -- replace with inorder successor and delete successor
        else:
            # find inorder successor
            successor = node.right
            successor_parent = node
            while successor.left:
                successor_parent = successor
                successor = successor.left
            # exchange the values
            node.val = successor.val
            successor.val = key
            # delete the successor --be careful that successor might have a right child
            if successor_parent.right == successor:
                successor_parent.right = successor.right
            else:
                successor_parent.left = successor.right
        return anchor.right

Approach 2:

    def deleteNode(self, root, key):
        # recursive
        if not root:
            return None
        if root.val > key:
            root.left = self.deleteNode(root.left, key)
            return root
        elif root.val < key:
            root.right = self.deleteNode(root.right, key)
            return root
        else:
            # this happens when root.val == key, need to delete the root
            if not root.left:
                return root.right
            if not root.right:
                return root.left
            # find successor
            node = root.right
            while node.left:
                node = node.left
            # the successor takes over the root's left subtree, and the new root is now root.right
            node.left = root.left
            return root.right

Code (C++)

Approach 1:

Approach 2: