450. Delete Node in a BST

450. Delete Node in a BST

https://www.youtube.com/watch?v=00r9qf7lgAk

https://www.youtube.com/watch?v=RDythl5S0fc
/**
 * 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 null;
      }
      
      if(key < root.val){
        root.left = deleteNode(root.left, key);
      }else if(key > root.val){
        root.right = deleteNode(root.right, key);
      }else{
        // we found the value to delete
        // three cases: both children are null, one of chidlren is null, neither is null
        if(root.left == null){ // this condition also covers the case when both children is null
          return root.right;
        }else if(root.right == null){
          return root.left;
        }else{ // neither is null. find the min from the right child
          TreeNode minNode = findMin(root.right);
          //get the min node value and give it to the root node
          root.val = minNode.val;
          // recursively delte the minnode 
          root.right = deleteNode(root.right, minNode.val);
        }
      }
      return root;
     
    }
    private TreeNode findMin(TreeNode root){
        while(root.left != null){
            root = root.left;
        }
        return root;
    }

}

 

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