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Formatted question description: https://leetcode.ca/all/450.html

# 450. Delete Node in a BST (Medium)

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

5
/ \
2   6
\   \
4   7


Companies:

Related Topics:
Tree

## Solution 1.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode deleteNode(TreeNode root, int key) {
if (root == null)
return root;
if (root.val == key) {
TreeNode dummyRoot = new TreeNode(0);
dummyRoot.left = root;
if (root.right != null) {
TreeNode successor = root.right;
if (successor.left == null) {
successor.left = root.left;
dummyRoot.left = successor;
} else {
TreeNode parent = successor, child = successor.left;
while (child.left != null) {
child = child.left;
parent = parent.left;
}
root.val = child.val;
parent.left = child.right;
}
} else
dummyRoot.left = root.left;
return dummyRoot.left;
} else {
TreeNode parent = root;
TreeNode child = root.val > key ? root.left : root.right;
while (child != null && child.val != key) {
parent = child;
child = child.val > key ? child.left : child.right;
}
if (child != null) {
boolean isLeftChild = parent.left == child;
if (child.left == null && child.right == null) {
if (isLeftChild)
parent.left = null;
else
parent.right = null;
} else if (child.left == null) {
if (isLeftChild)
parent.left = child.right;
else
parent.right = child.right;
} else if (child.right == null) {
if (isLeftChild)
parent.left = child.left;
else
parent.right = child.left;
} else {
TreeNode temp1 = child.right;
if (temp1.left == null) {
child.val = temp1.val;
child.right = temp1.right;
} else {
TreeNode temp2 = temp1.left;
while (temp2.left != null) {
temp2 = temp2.left;
temp1 = temp1.left;
}
child.val = temp2.val;
temp1.left = temp2.right;
}
}
}
return root;
}
}
}

############

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public TreeNode deleteNode(TreeNode root, int key) {
if (root == null) {
return null;
}
if (root.val > key) {
root.left = deleteNode(root.left, key);
return root;
}
if (root.val < key) {
root.right = deleteNode(root.right, key);
return root;
}
if (root.left == null) {
return root.right;
}
if (root.right == null) {
return root.left;
}
TreeNode node = root.right;
while (node.left != null) {
node = node.left;
}
node.left = root.left;
root = root.right;
return root;
}
}

• // OJ: https://leetcode.com/problems/delete-node-in-a-bst/
// Time: O(H)
// Space: O(H)
class Solution {
public:
TreeNode* deleteNode(TreeNode* root, int key) {
if (!root) return NULL;
if (root->val > key) root->left = deleteNode(root->left, key);
else if (root->val < key) root->right = deleteNode(root->right, key);
else if (root->left) {
auto p = root->left;
while (p->right) p = p->right;
root->val = p->val;
root->left = deleteNode(root->left, root->val);
} else if (root->right) {
auto p = root->right;
while (p->left) p = p->left;
root->val = p->val;
root->right = deleteNode(root->right, root->val);
} else {
delete root;
root = NULL;
}
return root;
}
};

• # 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: Optional[TreeNode], key: int) -> Optional[TreeNode]:
if root is None:
return None
if root.val > key:
root.left = self.deleteNode(root.left, key)
return root
if root.val < key:
root.right = self.deleteNode(root.right, key)
return root
if root.left is None:
return root.right
if root.right is None:
return root.left
node = root.right
while node.left:
node = node.left
node.left = root.left
root = root.right
return root

############

# 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
"""

def delete(root, pre):
if root.right:
p = root.right
while p.left:
p = p.left
p.left = root.left
if root is pre.left:
pre.left = root.right or root.left
if root is pre.right:
pre.right = root.right or root.left
root.left = None

if not root:
return root
pre = dummy = TreeNode(float("inf"))
dummy.left = root
p = dummy
while p:
if key > p.val:
pre = p
p = p.right
elif key < p.val:
pre = p
p = p.left
else:
delete(p, pre)
break
return dummy.left


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func deleteNode(root *TreeNode, key int) *TreeNode {
if root == nil {
return nil
}
if root.Val > key {
root.Left = deleteNode(root.Left, key)
return root
}
if root.Val < key {
root.Right = deleteNode(root.Right, key)
return root
}
if root.Left == nil {
return root.Right
}
if root.Right == nil {
return root.Left
}
node := root.Right
for node.Left != nil {
node = node.Left
}
node.Left = root.Left
root = root.Right
return root
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function deleteNode(root: TreeNode | null, key: number): TreeNode | null {
if (root == null) {
return root;
}
const { val, left, right } = root;
if (val > key) {
root.left = deleteNode(left, key);
} else if (val < key) {
root.right = deleteNode(right, key);
} else {
if (left == null && right == null) {
root = null;
} else if (left == null || right == null) {
root = left || right;
} else {
if (right.left == null) {
right.left = left;
root = right;
} else {
let minPreNode = right;
while (minPreNode.left.left != null) {
minPreNode = minPreNode.left;
}
const minVal = minPreNode.left.val;
root.val = minVal;
minPreNode.left = deleteNode(minPreNode.left, minVal);
}
}
}
return root;
}


• // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
let node = root.as_ref().unwrap().borrow();
if node.left.is_none() {
return node.val;
}
Self::dfs(&node.left)
}

pub fn delete_node(
mut root: Option<Rc<RefCell<TreeNode>>>,
key: i32,
) -> Option<Rc<RefCell<TreeNode>>> {
if root.is_some() {
let mut node = root.as_mut().unwrap().borrow_mut();
match node.val.cmp(&key) {
std::cmp::Ordering::Less => {
node.right = Self::delete_node(node.right.take(), key);
}
std::cmp::Ordering::Greater => {
node.left = Self::delete_node(node.left.take(), key);
}
std::cmp::Ordering::Equal => {
match (node.left.is_some(), node.right.is_some()) {
(false, false) => return None,
(true, false) => return node.left.take(),
(false, true) => return node.right.take(),
(true, true) => {
if node.right.as_ref().unwrap().borrow().left.is_none() {
let mut r = node.right.take();
r.as_mut().unwrap().borrow_mut().left = node.left.take();
return r;
} else {
let val = Self::dfs(&node.right);
node.val = val;
node.right = Self::delete_node(node.right.take(), val);
}
}
};
}
}
}
root
}
}