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.
Follow up: Can you solve it with time complexity O(height of tree)?
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. 
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.
Input: root = [], key = 0 Output: []
- The number of nodes in the tree is in the range
[0, 104]. -105 <= Node.val <= 105- Each node has a unique value.
rootis a valid binary search tree.-105 <= key <= 105
# 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 = rightclassSolution: defdeleteNode(self, root: TreeNode, key: int) ->TreeNode: ifrootisNone: returnNoneifroot.val==key: root=self.merge(root.left, root.right) elifroot.val>key: root.left=self.deleteNode(root.left, key) else: root.right=self.deleteNode(root.right, key) returnrootdefmerge(self, left: TreeNode, right: TreeNode) ->TreeNode: ifleftisNone: returnrightcurr=leftwhilecurr.rightisnotNone: curr=curr.rightcurr.right=rightreturnleft# 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}defdelete_node(root,key)returnnilifroot.nil?ifroot.val == keyroot=merge(root.left,root.right)elsifroot.val > keyroot.left=delete_node(root.left,key)elseroot.right=delete_node(root.right,key)endrootend# @param {TreeNode} left# @param {TreeNode} right# @return {TreeNode}defmerge(left,right)returnrightifleft.nil?curr=leftcurr=curr.rightuntilcurr.right.nil?curr.right=rightleftend