Search in a binary search tree

这题没难度，首先看下题目：

Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL.

For example, 

Given the tree:
        4
       / \
      2   7
     / \
    1   3

And the value to search: 2
You should return this subtree:

      2     
     / \   
    1   3
In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.

Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as [], not null.

知道BST的定义就知道该怎么做了，某个节点的左子树上的叶子节点的值都小于该节点的值，右子树反之。知道这个定义，就很容易写出算法来：

class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        if (root->val == val) {
            return root;
        }
        if (nullptr != root->left && val < root->val) {
            return searchBST(root->left, val);
        } else if (nullptr != root->right && val > root->val) {
            return searchBST(root->right, val);
        }
        return nullptr;
    }
};