Lintcode: Search Range in Binary Search Tree

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

Tags Expand 

Binary Search Tree Binary Tree

Reference: 

http://ryanleetcode.blogspot.com/2015/04/search-range-in-binary-search-tree.html //codes
http://www.code123.cc/docs/leetcode-notes/binary_search_tree/search_range_in_binary_search_tree.html    //thoughts

The essence is inorder traverses of BST
For each node,

its value is v

if v > k1, visit left subtree recursively

if k1 <= v <= k2, store the value

if v < k2, visit right subtree recursively

Time: O(n)

Space: O(L)

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/** * 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> result; searchRangeHelper(root, k1, k2, result); return result; } void searchRangeHelper(TreeNode* root, int k1, int k2, vector<int>& res){ if(root == NULL) return; if(root->val > k1) searchRangeHelper(root->left, k1, k2, res); if(root->val >= k1 && root->val <= k2) res.push_back(root->val); if(root->val < k2) searchRangeHelper(root->right, k1, k2, res); } };