[Leetcode]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:

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

思路：分两种情况，一种是要删除的节点只有一个孩子，另一种是要删除的节点有两个孩子；
如果只有一个孩子，那么我们让这个节点引用到他非空的孩子上去,即root = root.right或
root = root.left;如果有两个孩子的话，那么有两种办法，一种是找他右边的最小值，另一种
是找他左边的的最大值。将值赋给该节点，然后删除右边最小值或左边最大值；

 /**
  * 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;
         else if (key<root.val)
             root.left = deleteNode(root.left,key);
         else if (key>root.val)
             root.right = deleteNode(root.right,key);
         else {
             if (root.left!=null&&root.right!=null){
                 TreeNode tmp = findMin(root.right);
                 root.val = tmp.val;
                 root.right = deleteNode(root.right,root.val);
             }
             else {
                 TreeNode tmp = root;
                 if (root.left==null)
                     root = root.right;
                 else if (root.right==null)
                     root = root.left;
                 tmp = null;
             }
         }
         return root;
     }
     private TreeNode findMin(TreeNode root){
         if (root==null)
             return null;
         if (root.left==null)
             return root;
         return findMin(root.left);
     }
 }