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   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

我们先来看一种递归的解法，首先判断根节点是否为空。由于BST的左<根<右的性质，使得我们可以快速定位到要删除的节点，我们对于当前节点值不等于key的情况，根据大小关系对其左右子节点分别调用递归函数。若当前节点就是要删除的节点，我们首先判断是否有一个子节点不存在，那么我们就将root指向另一个节点，如果左右子节点都不存在，那么root就赋值为空了，也正确。难点就在于处理左右子节点都存在的情况，我们需要在右子树找到最小值，即右子树中最左下方的节点，然后将该最小值赋值给root，然后再在右子树中调用递归函数来删除这个值最小的节点，参见代码如下：

/** * 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 root;        }else if(key<root.val){            root.left = deleteNode(root.left, key); //要删除的点在左边，返回左节点        }else if(key>root.val){            root.right = deleteNode(root.right, key);        }else{            if(root.left!=null && root.right != null){                root.val = findMin(root.right).val;                root.right = deleteNode(root.right, root.val); //要记得删除移上来的那一点            }else{                if(root.left==null)                    root=root.right;                else root=root.left;            }        }        return root;            }        public TreeNode findMin(TreeNode root){        while(root.left != null){            root = root.left;        }        return root;    }}