> For the complete documentation index, see [llms.txt](https://junnie.gitbook.io/nine-chapter/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://junnie.gitbook.io/nine-chapter/7.two-pointers/653.-two-sum-iv-input-is-a-bst-m.md).

# 653. Two Sum IV - Input is a BST (M)

Given the `root` of a Binary Search Tree and a target number `k`, return *`true` if there exist two elements in the BST such that their sum is equal to the given target*.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/09/21/sum_tree_1.jpg)

```
Input: root = [5,3,6,2,4,null,7], k = 9
Output: true
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/09/21/sum_tree_2.jpg)

```
Input: root = [5,3,6,2,4,null,7], k = 28
Output: false
```

 

**Constraints:**

* The number of nodes in the tree is in the range `[1, 104]`.
* `-104 <= Node.val <= 104`
* `root` is guaranteed to be a **valid** binary search tree.
* `-105 <= k <= 105`

### Solution:

Solution 1: hashset + DFS, Time Complexity:`O(n)`, Space Complexity:`O(n)`

```
    Set<Integer> hashset = new HashSet<>();
    public boolean findTarget(TreeNode root, int k) {
        if (root == null) {
            return false;
        }
        if (hashset.contains(k - root.val)) {
            return true;
        }
        hashset.add(root.val);
        return findTarget(root.left, k) || findTarget(root.right, k);
    }
```

Solution 2: inorder traversal得到递增的array，two pointers. Time Complexity:`O(n)`, Space Complexity:`O(n)`

```
    public boolean findTarget(TreeNode root, int k) {
        List<Integer> nums = new ArrayList<>();
        inorder(root, nums);
        for(int i = 0, j = nums.size()-1; i<j;){
            if(nums.get(i) + nums.get(j) == k)return true;
            if(nums.get(i) + nums.get(j) < k)i++;
            else j--;
        }
        return false;
    }

    public void inorder(TreeNode root, List<Integer> nums){
        if(root == null)return;
        inorder(root.left, nums);
        nums.add(root.val);
        inorder(root.right, nums);
    }
```