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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:

  1. Search for a node to remove.
  2. 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

  • /**
     * 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;
        }
    }
    
  • /**
     * 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;
        }
    };
    
  • # 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.
     * 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
        }
    }
    
    

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