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   7Given 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       7Another 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;        }    }}