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

在top solution里面有个很好的解释：

Steps:

1. Recursively find the node that has the same value as the key, while setting the left/right nodes equal to the returned subtree
2. Once the node is found, have to handle the below 4 cases

• node doesn't have left or right - return null
• node only has left subtree- return the left subtree
• node only has right subtree- return the right subtree
• node has both left and right - find the minimum value in the right subtree, set that value to the currently found node, then recursively delete the minimum value in the right subtree

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