Leetcode/Uber -- 450. Delete Node in a BST(dfs)

450. Delete Node in a BST(dfs)

• Difficulty: Medium

https://leetcode.com/problems/delete-node-in-a-bst/

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.

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

 思路: recursion

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

• node doesn't have left or right - return null
• node only has left subtree- return the left subtree
• node only has right subtree- return the right subtree
• 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

Complexity:O(lgN -- height of tree).

public class Solution {
    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 = findMin(root.right);
            root.val=minNode.val;
            root.right=deleteNode(root.right,minNode.val);
        }
        return root;   
    }
    public TreeNode findMin(TreeNode root){
        if(root.left==null)return root;
        return findMin(root.left);
    }
}