题目描述：
Given two values k1 and k2 (where k1 < k2) and a root pointer to a Binary Search Tree. Find all the keys of tree in range k1 to k2. i.e. print all x such that k1<=x<=k2 and x is a key of given BST. Return all the keys in ascending order.
Have you met this question in a real interview? Yes ExampleIf k1 = 10 and k2 = 22, then your function should return[12, 20, 22].
20 / \ 8 22 / \ 4 12 题目思路：因为是bst，所以如果root->val > K2，所有右子树的值都不会在范围内，答案为左子树的搜索答案；root->val < K1也是同理。当root->val在范围内时，我们可以先得到左子树和右子树的答案，然后放入最终的答案vector中（因为是bst，所以只要按照left->root->right的顺序写就一定是排序好了的）。
Mycode（AC = 54ms）：
/** * 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: /** * @param root: The root of the binary search tree. * @param k1 and k2: range k1 to k2. * @return: Return all keys that k1<=key<=k2 in ascending order. */ vector<int> searchRange(TreeNode* root, int k1, int k2) { // write your code here vector<int> ans, left, right; if (root == NULL) { return ans; } // don't need to consider left tree else if (root->val < k1) { ans = searchRange(root->right, k1, k2); } // don't need to consider right tree else if (root->val > k2) { ans = searchRange(root->left, k1, k2); } else { // get answers of left tree and right tree left = searchRange(root->left, k1, k2); right = searchRange(root->right, k1, k2); // write answers into results, the order // should be left->root->right for (int i = 0; i < left.size(); i++) { ans.push_back(left[i]); } ans.push_back(root->val); for (int i = 0; i < right.size(); i++) { ans.push_back(right[i]); } } return ans; } };