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

Example 1:

Input:

    5
   / \
  3   6
 / \   \
2   4   7

Target = 9


Output:
 True

Example 2:

Input:

    5
   / \
  3   6
 / \   \
2   4   7

Target = 28


Output:
 False

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);
    }

Solution 3: The idea is to use binary search method. For each node, we check ifk - node.valexists in this BST.

Time Complexity:O(nlogn), Space Complexity:O(height).

    public boolean findTarget(TreeNode root, int k) {
        return dfs(root, root,  k);
    }

    public boolean dfs(TreeNode root,  TreeNode cur, int k){
        if(cur == null)return false;
        return search(root, cur, k - cur.val) || dfs(root, cur.left, k) || dfs(root, cur.right, k);
    }

    public boolean search(TreeNode root, TreeNode cur, int value){
        if(root == null)return false;
        return (root.val == value) && (root != cur) 
            || (root.val < value) && search(root.right, cur, value) 
                || (root.val > value) && search(root.left, cur, value);
    }

.