Question

Source

• lintcode: (11) Search Range in Binary Search Tree

Problem

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] .

  20
   /  \
  8   22
 / \
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++

/**
 * 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) {
    vector<int> result;
    inorder_dfs(result, root, k1, k2);

    return result;
  }

private:
  void inorder_dfs(vector<int> &ret, TreeNode *root, int k1, int k2) {
    if (NULL == root) {
      return;
    }

    inorder_dfs(ret, root->left, k1, k2);
    if ((root->val >= k1) && (root->val <= k2)) {
      ret.push_back(root->val);
    }
    inorder_dfs(ret, root->right, k1, k2);
  }
};

Java

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *   public int val;
 *   public TreeNode left, right;
 *   public TreeNode(int val) {
 *     this.val = val;
 *     this.left = this.right = null;
 *   }
 * }
 */
public class Solution {
  /**
   * @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.
   */
  public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
    ArrayList<Integer> result = new ArrayList<Integer>();
    helper(root, k1, k2, result);

    return result;
  }

  private void helper(TreeNode root, int k1, int k2, ArrayList<Integer> result) {
    if (root == null) return;

    // in-order binary tree iteration
    helper(root.left, k1, k2, result);
    if (k1 <= root.val && root.val <= k2) {
      result.add(root.val);
    }
    helper(root.right, k1, k2, result);
  }
}

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：

  void inorder_dfs(vector<int> &ret, TreeNode *root, int k1, int k2) {
    if (NULL == root) {
      return;
    }

    if ((NULL != root->left) && (root->val > k1)) {
      inorder_dfs(ret, root->left, k1, k2);
    } // cut-off for left sub tree

    if ((root->val >= k1) && (root->val <= k2)) {
      ret.push_back(root->val);
    } // add valid value

    if ((NULL != root->right) && (root->val < k2)) {
      inorder_dfs(ret, root->right, k1, k2);
    } // cut-off for right sub tree
  }

「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。