leetcode669

二叉树的题目无脑递归做，关键还是在三板斧——确定参数和返回值、终止条件和单层逻辑体

public TreeNode trimBST(TreeNode root, int low, int high) {
    if(root == null) return null;
    if(root.val < low) {
        if(root.right != null) return trimBST(root.right, low, high);
        else return null;
    }
    if(root.val > high) {
        if(root.left != null) return trimBST(root.left, low, high);
        else return null;
    }
    root.left = trimBST(root.left, low, high);
    root.right = trimBST(root.right, low, high);
    return root;
}