LintCode : Remove Node in Binary Search Tree

Description:

Given 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.

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

/**
 * 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.right == null) {
                return root.left;
            }
            if (root.left == null) {
                return root.right;
            }
            TreeNode t = root;
            root = min(t.right);
            root.right = deleteMin(t.right);
            root.left = t.left;
        }
        return root;
    }
    public TreeNode deleteMin(TreeNode x) {
        if (x.left == null) {
            return x.right;
        }
        x.left = deleteMin(x.left);
        return x;
    }
    public TreeNode min(TreeNode x) {
        if (x.left == null) {
            return x;
        }
        return min(x.left);
    }
}

 

Please refer to my post here : http://www.cnblogs.com/dingjunnan/p/5417784.html