LiteCode-Search Range in Binary Search Tree_litecode刷题-优快云博客

本文介绍了一种在二叉搜索树中查找指定区间内所有元素的算法，通过广度优先搜索实现，确保了返回结果的升序排列。适用于需要在二叉树中快速查找特定范围元素的应用场景。

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Description

Given a binary search tree and a range [k1, k2], return all elements in the given range.

Example

If k1 = 10 and k2 = 22, then your function should return [12, 20, 22].
    20
   /  \
  8   22
 / \
4   12

题目模板

/**
* 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: param root: The root of the binary search tree
   * @param k1: An integer
   * @param k2: An integer
   * @return: return: Return all keys that k1<=key<=k2 in ascending order
   */
  vector<int> searchRange(TreeNode * root, int k1, int k2) {
      // write your code here
  }
};

题目大意

给你一个二叉搜索树，和两个数k1，k2，让你输出在数中值介于这两个数的数。

大概思路

没啥好说的，遍历书吧，深搜广搜都可以。

/**
* 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: param root: The root of the binary search tree
   * @param k1: An integer
   * @param k2: An integer
   * @return: return: Return all keys that k1<=key<=k2 in ascending order
   */
  vector<int> searchRange(TreeNode * root, int k1, int k2) {
      // write your code here
      queue<TreeNode*> q;
      vector<int> v;
      if(!root)
        return v;

      q.push(root);
      while(!q.empty()){
        TreeNode *t = q.front();
        q.pop();
        if(t->val >= k1 && t->val <= k2)
          v.push_back(t->val);
        if(t->left)
          q.push(t->left);
        if(t->right)
          q.push(t->right);
      }
      return v;
  }
};

细节方面

注意树可能为空直接返回空vector就可以了。
题目链接: https://www.lintcode.com/problem/search-range-in-binary-search-tree/description