LeetCode 450. Delete Node in a BST

原题网址：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

方法：根据情况进行处理。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null) return null;
        if (key < root.val) {
            root.left = deleteNode(root.left, key);
            return root;
        }
        if (key > root.val) {
            root.right = deleteNode(root.right, key);
            return root;
        }
        if (root.left != null) {
            TreeNode left = root.left;
            if (left.right == null) {
                left.right = root.right;
                return left;
            }
            while (left.right != null && left.right.right != null) {
                left = left.right;
            }
            if (left.right.left == null) {
                root.val = left.right.val;
                left.right = null;
                return root;
            }
            root.val = left.right.val;
            left.right = left.right.left;
            return root;
        } else {
            return root.right;
        }
    }
}