[LeetCode]Delete Node in a BST

Question
Delete a node in a BST.

Basically, the deletion can be divided into two stages:

Search for a node to remove.
If the node is found, delete the node.
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

本题难度Medium。

递归

【复杂度】
时间 O(logN) 空间 O(1)

【思路】
利用递归法查找key，找到该node后需要讨论：

1. node没有儿子。返回null
2. node有1个儿子。返回儿子
3. node有2个儿子。这个比较麻烦。我们利用node的右子树的最小值替换node的值，然后到node的右子树去删除该最小值即可。

【代码】

/**
 * 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) {
        //require
        if(root==null)
            return root;
        //invariant
        if(key<root.val)
            root.left=deleteNode(root.left,key);
        else if(root.val<key)
            root.right=deleteNode(root.right,key);
        else{
            //two children
            if(root.left!=null&&root.right!=null){
                TreeNode tmp=findMin(root.right);
                root.val=tmp.val;
                root.right=deleteNode(root.right,root.val);
            }else{
                //one child or zero
                if(root.left==null)
                    root=root.right;
                else
                    root=root.left;
            }
        }
        //ensure
        return root;
    }
    private TreeNode findMin(TreeNode root){
        while(root.left!=null)
            root=root.left;
        return root;
    }
}

参考
《数据结构与算法分析：C语言描述》 第2版 4.3.5节