问题：

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   7Target = 9Output: True

Example 2:

Input:     5   / \  3   6 / \   \2   4   7Target = 28Output: False

解决：

① 中序遍历二叉搜索树可以得到一个递增数组，只要查找该数组中是否包含两个数之和为target。

/**

 * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode(int x) { val = x; }  * }  */ public class Solution { // 33ms     public boolean findTarget(TreeNode root, int k) {         if(root == null) return false;         List<Integer> list = new ArrayList<>();         inorder(root,list);         int head = 0;         int tail = list.size() - 1;         while(head < tail){             if(list.get(head) + list.get(tail) == k){                 return true;             }else if(list.get(head) + list.get(tail) < k){                 head ++;             }else{                 tail --;             }         }         return false;     }     public void inorder(TreeNode node,List<Integer> list){         if(node == null) return;         inorder(node.left,list);         list.add(node.val);         inorder(node.right,list);     } }

② 先序遍历二叉搜索树，根据遍历到的节点再查找target- node.val即可。

public class Solution { //23ms

    public boolean findTarget(TreeNode root, int k) {         return helper(root, root, k);     }     public boolean helper(TreeNode root, TreeNode curNode, int k) {         if (curNode == null) return false;         return preorder(root, curNode, k - curNode.val) || helper(root, curNode.left, k) || helper(root, curNode.right, k);     }     public boolean preorder(TreeNode root, TreeNode curNode, int k) {         if (root == null) return false;         if (root != curNode && root.val == k) return true;         return (root.val < k) ? preorder(root.right, curNode, k) : preorder(root.left, curNode, k);     } }