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.

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

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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: []

Solutions

TreeNode

// 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;
		}
}

Solution 1

/**
 * Time complexity :
 * Space complexity :
 */

class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if(root == null) {
            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== null && root.right == null) { // 0 children
                root = null;
            } else if(root.left == null) { // 1 children, right side
                root = root.right;
            } else if(root.right == null) { // 1 children, left side
                root = root.left;
            } else { // 2 children
                TreeNode rightMinNode = findMinNode(root.right);
                root.val = rightMinNode.val;
                root.right = deleteNode(root.right, rightMinNode.val);
            }
        }
        return root;
    }
    
    private TreeNode findMinNode(TreeNode node) {
        return (node.left == null)? node: findMinNode(node.left);
    }
}

Follow up

• Can you solve it with time complexity O(height of tree)?