#11 Search Range in Binary Search Tree

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

Example

If 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;    }};