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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.

Follow up: Can you solve it with time complexity O(height of tree)?

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

Solutions (Python)

1. DFS

# 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 deleteNode(self, root: TreeNode, key: int) -> TreeNode:
        if root is None:
            return None

        if root.val == key:
            root = self.merge(root.left, root.right)
        elif root.val > key:
            root.left = self.deleteNode(root.left, key)
        else:
            root.right = self.deleteNode(root.right, key)

        return root

    def merge(self, left: TreeNode, right: TreeNode) -> TreeNode:
        if left is None:
            return right

        curr = left
        while curr.right is not None:
            curr = curr.right
        curr.right = right

        return left

Solutions (Ruby)

1. DFS

# Definition for a binary tree node.
# class TreeNode
#     attr_accessor :val, :left, :right
#     def initialize(val = 0, left = nil, right = nil)
#         @val = val
#         @left = left
#         @right = right
#     end
# end
# @param {TreeNode} root
# @param {Integer} key
# @return {TreeNode}
def delete_node(root, key)
  return nil if root.nil?

  if root.val == key
    root = merge(root.left, root.right)
  elsif root.val > key
    root.left = delete_node(root.left, key)
  else
    root.right = delete_node(root.right, key)
  end

  root
end

# @param {TreeNode} left
# @param {TreeNode} right
# @return {TreeNode}
def merge(left, right)
  return right if left.nil?

  curr = left
  curr = curr.right until curr.right.nil?
  curr.right = right

  left
end