給定兩個值 k1 和 k2（k1 < k2）和一個二叉查詢樹的根節點。找到樹中所有值在 k1 到 k2 範圍內的節點。即列印所有x (k1 <= x <= k2) 其中 x 是二叉查詢樹的中的節點值。返回所有升序的節點值。

如果有 k1 = 10 和 k2 = 22, 你的程式應該返回 [12, 20, 22].

    20
   /  \
  8   22
 / \
4   12

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:

    void middle_order(TreeNode *root,vector<int> &base,int k1,int k2){
        if(!root)
            return ;
        middle_order(root->left,base,k1,k2);
        if(root->val<=k2&&root->val>=k1)
            base.push_back(root->val);
        middle_order(root->right,base,k1,k2);
    }
    
    vector<int> searchRange(TreeNode* root, int k1, int k2) {
        vector<int> ret;
        if(!root)
            return ret;
        middle_order(root,ret,k1,k2);
        return ret;
    }
};