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:

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

Example 2:

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);
    }