700. Search in a Binary Search Tree

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.

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给定一个二叉搜索树，要求从树中搜索一个等于val的节点。二叉搜索树是左子树节点小于根节点，右子树节点大于根节点的树。所以搜索时只需要和根节点比较大小，递归的在左子树或右子树中搜索即可。

完整代码如下：

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

本代码提交AC，用时52MS