Search Range in Binary Search Tree

Question

Problem Statement

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.

Example

If k1 = 10 and k2 = 22, then your function should return [12, 20, 22].

  1.     20
  2.    /  \
  3.   8   22
  4.  / \
  5. 4   12

题解 - 中序遍历

中等偏易难度题，本题涉及到二叉查找树的按序输出，应马上联想到二叉树的中序遍历，对于二叉查找树而言，使用中序遍历即可得到有序元素。对每次访问的元素加以判断即可得最后结果，由于 OJ 上给的模板不适合递归处理，新建一个私有方法即可。

C++

  1. /**
  2.  * Definition of TreeNode:
  3.  * class TreeNode {
  4.  * public:
  5.  *     int val;
  6.  *     TreeNode *left, *right;
  7.  *     TreeNode(int val) {
  8.  *         this->val = val;
  9.  *         this->left = this->right = NULL;
  10.  *     }
  11.  * }
  12.  */
  13. class Solution {
  14. public:
  15.     /**
  16.      * @param root: The root of the binary search tree.
  17.      * @param k1 and k2: range k1 to k2.
  18.      * @return: Return all keys that k1<=key<=k2 in ascending order.
  19.      */
  20.     vector<int> searchRange(TreeNode* root, int k1, int k2) {
  21.         vector<int> result;
  22.         inorder_dfs(result, root, k1, k2);
  24.         return result;
  25.     }
  27. private:
  28.     void inorder_dfs(vector<int> &ret, TreeNode *root, int k1, int k2) {
  29.         if (NULL == root) {
  30.             return;
  31.         }
  33.         inorder_dfs(ret, root->left, k1, k2);
  34.         if ((root->val >= k1) && (root->val <= k2)) {
  35.             ret.push_back(root->val);
  36.         }
  37.         inorder_dfs(ret, root->right, k1, k2);
  38.     }
  39. };

Java

  1. /**
  2.  * Definition of TreeNode:
  3.  * public class TreeNode {
  4.  *     public int val;
  5.  *     public TreeNode left, right;
  6.  *     public TreeNode(int val) {
  7.  *         this.val = val;
  8.  *         this.left = this.right = null;
  9.  *     }
  10.  * }
  11.  */
  12. public class Solution {
  13.     /**
  14.      * @param root: The root of the binary search tree.
  15.      * @param k1 and k2: range k1 to k2.
  16.      * @return: Return all keys that k1<=key<=k2 in ascending order.
  17.      */
  18.     public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
  19.         ArrayList<Integer> result = new ArrayList<Integer>();
  20.         helper(root, k1, k2, result);
  22.         return result;
  23.     }
  25.     private void helper(TreeNode root, int k1, int k2, ArrayList<Integer> result) {
  26.         if (root == null) return;
  28.         // in-order binary tree iteration
  29.         helper(root.left, k1, k2, result);
  30.         if (k1 <= root.val && root.val <= k2) {
  31.             result.add(root.val);
  32.         }
  33.         helper(root.right, k1, k2, result);
  34.     }
  35. }

源码分析

以上为题解思路的简易实现，可以优化的地方为「剪枝过程」的处理——不递归遍历不可能有解的节点。优化后的inorder_dfs如下：

  1.     void inorder_dfs(vector<int> &ret, TreeNode *root, int k1, int k2) {
  2.         if (NULL == root) {
  3.             return;
  4.         }
  6.         if ((NULL != root->left) && (root->val > k1)) {
  7.             inorder_dfs(ret, root->left, k1, k2);
  8.         } // cut-off for left sub tree
  10.         if ((root->val >= k1) && (root->val <= k2)) {
  11.             ret.push_back(root->val);
  12.         } // add valid value
  14.         if ((NULL != root->right) && (root->val < k2)) {
  15.             inorder_dfs(ret, root->right, k1, k2);
  16.         } // cut-off for right sub tree
  17.     }

Warning 「剪枝」的判断条件容易出错，应将当前节点的值与k1和k2进行比较而不是其左子节点或右子节点的值。