450. Delete Node in a BST

Problem

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, 10^4].

• -10^5 <= Node.val <= 10^5

• Each node has a unique value.

• root is a valid binary search tree.

• -10^5 <= key <= 10^5

  Follow up: Could you solve it with time complexity O(height of tree)?

Solution (Java)

/**
 * 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 {
    /*
     * Steps:
     * 1. Recursively find the node that has the same value as the key, while setting the left/right nodes equal
     * to the returned subtree
     * 2. Once the node is found, have to handle the below 4 cases
     * a. node doesn't have left or right - return null
     * b. node only has left subtree- return the left subtree
     * c. node only has right subtree- return the right subtree
     * d. node has both left and right - find the minimum value in the right subtree, set that value
     * to the currently found node, then recursively delete the minimum value in the right subtree
     */
    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null) {
            return root;
        }
        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 == null) {
                return root.right;
            } else if (root.right == null) {
                return root.left;
            }
            TreeNode minNode = getMin(root.right);
            root.val = minNode.val;
            root.right = deleteNode(root.right, root.val);
        }
        return root;
    }

    private TreeNode getMin(TreeNode node) {
        while (node.left != null) {
            node = node.left;
        }
        return node;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).