Lintcode: Remove Node in Binary Search Tree

iven a root of Binary Search Tree with unique value for each node.  Remove the node with given value. If there is no such a node with given value in the binary search tree, do nothing. You should keep the tree still a binary search tree after removal.

Have you met this question in a real interview? Yes
Example
Given binary search tree:

          5

       /    \

    3          6

 /    \

2       4

Remove 3, you can either return:

          5

       /    \

    2          6

      \

         4

or :

          5

       /    \

    4          6

 /   

2

分析：假设当前node 为root

1. if value < root.val, 要被删除的节点在左子树，往左子树递归，并把操作结束后的返回值作为新的root.left

2. if value > root.val, 要被删除的节点在右子树，往右子树递归, 并把操作结束后的返回值作为新的root.right

3. if root == null, 递归到了一个null点，说明要删的value不存在，return null，而这个null点的parent的相应子树本来也是null，对树的结构没有任何影响

4. if value == root.val，说明root是该被删除的了

　　A. if root.left == null, return root.right

　　B. if root.right == null, return root.left(这两个case其实包含了只有一个child和一个child都没有的三种情况)

　　C. 如果两个children都存在，从右子树中找最小的node，与root交换，再递归调用函数在右子树中删除root.val

 /**
  * Definition of TreeNode:
  * public class TreeNode {
  *     public int val;
  *     public TreeNode left, right;
  *     public TreeNode(int val) {
  *         this.val = val;
  *         this.left = this.right = null;
  *     }
  * }
  */
 public class Solution {
     /**
      * @param root: The root of the binary search tree.
      * @param value: Remove the node with given value.
      * @return: The root of the binary search tree after removal.
      */
     public TreeNode removeNode(TreeNode root, int value) {
         // write your code here
         if (root == null) return null;
         if (value < root.val)
             root.left = removeNode(root.left, value);
         else if (value > root.val)
             root.right = removeNode(root.right, value);
         else {
             if (root.left == null) return root.right;
             if (root.right == null) return root.left;
             TreeNode minOfRight = findMin(root.right);
             //swap root and minOfRight
             int temp = root.val;
             root.val = minOfRight.val;
             minOfRight.val = temp;
             root.right = removeNode(root.right, minOfRight.val);
         }
         return root;
     }

     public TreeNode findMin(TreeNode cur) {
         while (cur.left != null) {
             cur = cur.left;
         }
         return cur;
     }
 }