Description
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.
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
Follow up: Could you solve it with time complexity O(height of tree)?
Solutions
Solution 1
PythonJavaC++GoTypeScriptRust
# 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(code-box)
/**
* 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;
}
}(code-box)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* deleteNode(TreeNode* root, int key) {
if (!root) return root;
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) return root->right;
if (!root->right) return root->left;
TreeNode* node = root->right;
while (node->left) node = node->left;
node->left = root->left;
root = root->right;
return root;
}
};(code-box)
/**
* 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
}(code-box)
/**
* 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;
}(code-box)
// 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::cell::RefCell;
use std::rc::Rc;
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
}
}(code-box)
