LeetCode.450 Delete Node in a BST（经典删除二叉树某个节点，必备题）

题目：

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
分析：/**
* 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) {
//给定规则二叉树，删除其中节点，调整原来二叉树使得保持原规则
//思路：找到要删除节点，之后用其右子树中最小的代替该值
if(root==null) return root;
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){
//双子树均存在的情况
root.val=findMin(root.right).val;
//删除右孩子中最小的节点
root.right=deleteNode(root.right,root.val);
}else{
//当前节点即为删除节点
root=(root.left!=null)?root.left:root.right;
}
return root;
}
public TreeNode findMin(TreeNode root){
//查找最左边的节点
if(root==null) return null;
else if(root.left==null){
return root;
}else{
return findMin(root.left);
}
}
}