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
    }
}

450 - Delete Node in a BST

450. Delete Node in a BST (Medium)

Solution 1.

All Problems

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450. Delete Node in a BST (Medium)

Solution 1.

All Problems

All Solutions