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.

class Solution {
    public TreeNode searchBST(TreeNode root, int val) {
        while(root != null && root.val != val){
            root = root.val > val ? root.left : root.right;
        }
        return root;
    }
}

我是sb？

class Solution {
    public TreeNode searchBST(TreeNode root, int val) {
            if(root == null || root.val == val) return root;
            return val < root.val? searchBST(root.left, val): searchBST(root.right, val);
        }
}

注意啊，上下是平行的关系，不是if-else的关系，否则会有nullpointerexception出现