LeetCode - Trim a Binary Search Tree

Given a binary search tree and the lowest and highest boundaries as L and R, trim the tree so that all its elements lies in [L, R] (R >= L). You might need to change the root of the tree, so the result should return the new root of the trimmed binary search tree.

Example 1:
Input: 
    1
   / \
  0   2

  L = 1
  R = 2

Output: 
    1
      \
       2
Example 2:
Input: 
    3
   / \
  0   4
   \
    2
   /
  1

  L = 1
  R = 3

Output: 
      3
     / 
   2   
  /
 1

　　正确方法其实应该是在遍历的过程中就修改二叉树，移除不合题意的结点。当然对于二叉树的题，十有八九都是要用递归来解的。首先判断如果root为空，那么直接返回空即可。然后就是要看根结点是否在范围内，如果根结点值小于L，那么返回对其右子结点调用递归函数的值；如果根结点大于R，那么返回对其左子结点调用递归函数的值。如果根结点在范围内，将其左子结点更新为对其左子结点调用递归函数的返回值，同样，将其右子结点更新为对其右子结点调用递归函数的返回值。最后返回root即可，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; }
 * }
 */
class Solution {
    public TreeNode trimBST(TreeNode root, int L, int R) {
        if(root == null){
            return null;
        }
        if(root.val > R){
            return trimBST(root.left, L, R);
        }
        else if(root.val < L){
            return trimBST(root.right, L, R);
        }
        root.left = trimBST(root.left, L, R);
        root.right = trimBST(root.right, L, R);
        return root;
        
    }
}