Lintcode 11 Search Range in Binary Search Tree

最新推荐文章于 2018-08-08 14:59:50 发布

转载最新推荐文章于 2018-08-08 14:59:50 发布 · 272 阅读

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方法。通过递归遍历树结构，确保找到的所有元素均按升序排列并位于指定范围内。此算法适用于需要高效查询特定范围数据的应用场景。

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].

public class Solution {
    private ArrayList<Integer> results;
    /**
     * @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 increasing order.
     */
    public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
        results = new ArrayList<Integer>();
        helper(root, k1, k2);
        return results;
    }
    
    private void helper(TreeNode root, int k1, int k2) {
        if (root == null) {
            return;
        }
        if (root.val > k1) {
            helper(root.left, k1, k2);
        }
        if (root.val >= k1 && root.val <= k2) {
            results.add(root.val);
        }
        if (root.val < k2) {
            helper(root.right, k1, k2);
        }
    }
}


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 increasing order.
     */
    vector<int> ans;
    void  dfs(TreeNode* node, int k1, int k2) {
       
        if (node->left && node->val>=k1)
            dfs(node->left, k1, k2);
         if (node->val >= k1 && node->val <= k2)
            ans.push_back(node->val);
        if (node->right && node->val<=k2)
            dfs(node->right, k1, k2);
    }
    vector<int> searchRange(TreeNode* root, int k1, int k2) {
        // write your code here
        if (root==NULL)
            return ans;
        dfs(root, k1, k2);
        return ans;
    }
};