# 653 Two Sum IV - Input is a BST

## 653. Two Sum IV - Input is a BST

## 1. Question

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
```

## 2. Implementation

(1) DFS

```java
class Solution {
    public boolean findTarget(TreeNode root, int k) {
        Set<Integer> set = new HashSet<>();

        return findTarget(root, k, set);
    }

    public boolean findTarget(TreeNode node, int target, Set<Integer> set) {
        if (node == null) {
            return false;
        }

        if (set.contains(target - node.val)) {
            return true;
        }
        set.add(node.val);
        return findTarget(node.left, target, set) || findTarget(node.right, target, set);
    }
}
```

(2) BFS

```java
class Solution {
    public boolean findTarget(TreeNode root, int k) {
        Set<Integer> set = new HashSet<>();

        if (root == null) {
            return false;
        }

        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);

        while (!queue.isEmpty()) {
            int size = queue.size();

            for (int i = 0; i < size; i++) {
                TreeNode curNode = queue.remove();

                if (set.contains(k - curNode.val)) {
                    return true;
                }
                set.add(curNode.val);

                if (curNode.left != null) {
                    queue.add(curNode.left);
                }

                if (curNode.right != null) {
                    queue.add(curNode.right);
                }
            }
        }
        return false;
    }
}
```

## 3. Time & Space Complexity

DFS: 时间复杂度: O(n), 空间复杂度: O(n)

BFS: 时间复杂度: O(n), 空间复杂度: O(w), w为一层中最多node的个数


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