LeetCode[450]Delete Node in a BST(Java)

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.

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      

Solution:

Recursive遞歸

首先查找到val=key節點的所在位置，在查找的過程中，利用遞歸重建BST。

查找到之後，需要将目前節點重新指派，然後重新建構右子樹，方式是将右子樹的最左葉結點的值付給目前節點，然後删除該最左葉結點。

/**
 * 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(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 temp = search(root.right);
            root.val = temp.val;
            root.right = deleteNode(root.right, temp.val);
        }
        return root;
    }
    public TreeNode search(TreeNode node){
        while(node.left != null){
            node = node.left;
        }
        return node;
    }
}