Welcome to Subscribe On Youtube
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:
- Search for a node to remove.
- 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
Companies:
Microsoft, Google, Amazon
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 } }