653. Two Sum IV

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

用了最蠢的办法把节点先全部保存起来，再循环找到和为k的点

之前一直在用树递归写，思路大致是每个节点上先将k减去当前节点的val，然后再遍历当前节点的左右子树，相当于在bst树上进行查找值为k-val的点，一旦找到就返回true，但是一直没写对，，再看看。

class Solution {
public:
    bool findTarget(struct TreeNode* root, int k)
    {
        vector<int> temp;
        int i,j;
        inorder(root,temp);
        for(i=0;i<temp.size();i++){
            for(j=i+1;j<temp.size();j++){
                if(temp[i]+temp[j]==k) return true;
            }
        }
        return false;
    }


    void inorder(struct TreeNode* root,vector<int>& temp)
    {
        if(root==NULL) return;
        temp.push_back(root->val);
        inorder(root->left,temp);
        inorder(root->right,temp);
    }
};