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
 方法一将BST中的每一个结点的值存储在数组中，由于我们已经在Leetcode371中实现有序数组中的计算，因此可以采用如下代码来实现： 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/*计算BST中对应的结点个数*/
int NodeNums(struct TreeNode* root) 
{
    if(NULL == root) 
        return 0;

    return NodeNums(root->left) + NodeNums(root->right) + 1;
}

/*将结点转换成数组*/
void BST2Array(struct TreeNode* root, int* array, int* pos) 
{
    if(NULL == root) 
        return;
    
    BST2Array(root->left, array, pos);
    array[(*pos)++] = root->val;
    BST2Array(root->right, array, pos);
}

bool findTarget(struct TreeNode* root, int k) 
{
    if(NULL == root) 
        return false;

    int nodeNums = NodeNums(root);
    int* nodes = (int*)malloc(sizeof(int)*nodeNums);
    int pos = 0;
    BST2Array(root, nodes, &pos);
    
    int low = 0, high = nodeNums-1;
    while(low < high) 
    {
        int temp = nodes[low] + nodes[high];
        if(k == temp) 
            return true;
        else if(k < temp) 
            high--;
        else low++;
    }
    /*释放申请的内存*/
    free(nodes);
    return false;
}