leetcode450Delete 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

程序代码：

/** * 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;        if(key < root.val){            root.left = deleteNode(root.left, key);        }else if(key > root.val){            root.right = deleteNode(root.right, key);        }else{            //fnided the key node            if(root.right == null)                return root.left;            if(root.left == null)                return root.right;            //return the minim node in the right subtree            TreeNode rightmin = findMinimNode(root.right);            root.val = rightmin.val;            root.right = deleteNode(root.right, rightmin.val);        }        return root;    }        private TreeNode findMinimNode(TreeNode root){        while(root.left != null){            root = root.left;        }        return root;    }}

运行：

有人说直接用被删除节点的右节点来替换被删除节点，然后再直接把被删除节点的左节点接到被删节点的右子树中。这样就可以不用在递归删除了。但是种种方式很明显会增加树的高度，对后续数据结构的维护是不利的，到时候还是需要再平衡树，但是针对这个问题是可以的。