450. 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

含义：现在有一个二叉搜索树，现在要让你删除一个节点，并且保证整个BST的性质不变

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11         
12 public TreeNode deleteNode(TreeNode root, int key) {
13     if(root == null){
14         return null;
15     }
16     if(key < root.val){
17         root.left = deleteNode(root.left, key);
18     }else if(key > root.val){
19         root.right = deleteNode(root.right, key);
20     }else{
21         if(root.left == null){
22             return root.right;
23         }else if(root.right == null){
24             return root.left;
25         }
26         //右子树中最小值挪到当前root上，才能保持二叉树性质不变
27         TreeNode minNode = findMin(root.right);
28         root.val = minNode.val;
29         root.right = deleteNode(root.right, root.val);
30     }
31     return root;
32 }
33 
34 private TreeNode findMin(TreeNode node){
35     while(node.left != null){
36         node = node.left;
37     }
38     return node;
39 }
40 }