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450. Delete Node in a BST
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
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/** * 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; } }
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/** * 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; } };
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# 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
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/** * 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 }
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/** * 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; }
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// 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 } }