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

Method dfs O(h) time and space.

• standard dfs to find the node with value == key.
• Once node is found, replace the node with its child if the node only has one child.
• If node has two children, set the current node as the right child, traverse to the rightmost child of its left subtree and assign the original root's left subtree there.

# 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 not root:
            return None
        if root.val == key:
            if not root.left:
                return root.right
            elif not root.right:
                return root.left
            left_node = root.left
            right_node = root.right
            root = right_node
            while right_node.left:
                right_node = right_node.left
            right_node.left = left_node                         
        elif root.val < key:
            root.right = self.deleteNode(root.right, key)
        elif root.val > key: 
            root.left = self.deleteNode(root.left, key)
        return root