Two Sum IV - Input is a BST

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

思路：用inorder stack traverse，那么就是O(N)，跟Two Sum一样；

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean findTarget(TreeNode root, int k) {
        if(root == null) {
            return false;
        }
        HashSet<Integer> set = new HashSet<Integer>();
        return inorder(root, set, k);
    }
    
    private boolean inorder(TreeNode root, HashSet<Integer> set, int target) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode node = root;
        while(node != null) {
            stack.push(node);
            node = node.left;
        }
        
        while(!stack.isEmpty()) {
            node = stack.pop();
            if(set.contains(target - node.val)) {
                return true;
            } else {
                set.add(node.val);
            }
            
            if(node.right != null) {
                node = node.right;
                while(node != null) {
                    stack.push(node);
                    node = node.left;
                }
            }
        }
        return false;
    }
}

DFS inorder traverse:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean findTarget(TreeNode root, int k) {
        if(root == null) {
            return false;
        }
        HashSet<Integer> set = new HashSet<Integer>();
        return inorder(root, set, k);
    }
    
    private boolean inorder(TreeNode root, HashSet<Integer> set, int target) {
        if(root == null) {
            return false;
        }
        if(set.contains(target - root.val)) {
            return true;
        }
        set.add(root.val);
        return inorder(root.left, set, target) || inorder(root.right, set, target);
    }
}