669. Trim a Binary Search Tree

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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:

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Input:
    1
   / \
  0   2
  L = 1
  R = 2
Output:
    1
      \
       2

Example 2:

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Input:
    3
   / \
  0   4
   \
    2
   /
  1
  L = 1
  R = 3
Output:
      3
     /
   2   
  /

解法1： Recursion

应用BST的性质，如果当前是一个leaf并且不在[L,R]范围内，那么就返回null。
如果不是leaf并且leaf的value在范围内，那么就对左右子树各trim
如果不是leaf而不在范围内，那么就either对左子数或者右子数trim就可以了。

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/**
 * 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.left == null && root.right == null) {
            if (root.val >= L && root.val <= R) {
                return root;
            } else {
                return null;
            }
        }
        int val = root.val;
        if (val >= L && val <= R) {
            root.left = trimBST(root.left, L, R);
            root.right = trimBST(root.right, L, R);
            return root;
        } else if (val < L) {
            return trimBST(root.right, L, R);
        } else {
            // val > R
            return trimBST(root.left, L, R);            
        }
    }
}