leetcode450. 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:

Search for a node to remove.
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

1. 叶节点
    5
   / \
  2   6
   \   \
    4   7 （删除4）
结果为：
    5
   / \
  2   6
       \
        7
        
2. 只有左子树或是只有右子树
    5
   / \
  3   6
 / \   \
2   4   7（删除6）
结果为
    5
   / \
  3   6
 / \   
2   4   

3. 既有左子树又有右子树
    6
   / \
  3   7
 / \   \
2   5   8 （删除6）
   /
  4
首先找到6的左子树中的最大值为5，将5填充到6的位置
    5
   / \
  3   7
 / \   \
2   5   8 （删除5）
   /
  4
接着递归的在左子树中删除5，此时5满足只有一个子树的场景，因此直接用子树替换即可
    5
   / \
  3   7
 / \   \
2   4   8 （删除5）

    public TreeNode deleteNode(TreeNode cur, int key) {
        if(cur == null) return null;
        else if(cur.val == key) {
            if(cur.left != null && cur.right != null) {
                TreeNode left = cur.left;
                while(left.right != null) {
                    left = left.right;
                }
                cur.val = left.val;
                cur.left = deleteNode(cur.left, left.val);
            }else if(cur.left != null) {
                return cur.left;
            }else if(cur.right != null){
                return cur.right;
            }else {
                return null;
            }
        }else if(cur.val > key) {
            cur.left = deleteNode(cur.left, key);
        }else {
            cur.right = deleteNode(cur.right, key);
        }
        return cur;
    }