669. 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

M1: recursion

如果root.val比R大，说明要trim掉右边的节点，要返回的subtree在root的左边，即递归到root.left；

如果root.val比L小，说明要trim掉左边的节点，要返回的subtree在root的右边，递归到root.right；

如果root.val在[L, R]，说明有可能左右两边都需要trim，分别递归左右节点

time: O(n), space: O(height)

/**
 * 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 root;
        }
        if(root.val > R) {
            return trimBST(root.left, L, R);
        }
        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;
    }
}