247-线段树的查询 II

对于一个数组，我们可以对其建立一棵 线段树, 每个结点存储一个额外的值 count 来代表这个结点所指代的数组区间内的元素个数. (数组中并不一定每个位置上都有元素)

实现一个 query 的方法，该方法接受三个参数 root, start 和 end, 分别代表线段树的根节点和需要查询的区间，找到数组中在区间[start, end]内的元素个数。

注意事项

It is much easier to understand this problem if you finished Segment Tree Buildand Segment Tree Query first.

样例

对于数组 [0, 空，2, 3], 对应的线段树为：



query(1, 1), return 0

query(1, 2), return 1

query(2, 3), return 2

query(0, 2), return 2

标签

二叉树 LintCode 版权所有 线段树

思路

与 lintcode-202-线段树的查询 类似，只需注意边界条件

code

/**

 * Definition of SegmentTreeNode:

 * class SegmentTreeNode {

 * public:

 *     int start, end, count;

 *     SegmentTreeNode *left, *right;

 *     SegmentTreeNode(int start, int end, int count) {

 *         this->start = start;

 *         this->end = end;

 *         this->count = count;

 *         this->left = this->right = NULL;

 *     }

 * }

 */

class Solution {

public:

    /*

     * @param root: The root of segment tree.

     * @param start: start value.

     * @param end: end value.

     * @return: The count number in the interval [start, end]

     */

    int query(SegmentTreeNode * root, int start, int end) {

        // write your code here

        if(root == NULL || start > end) {

            return 0;

        }

        int mid = (root->start + root->end) / 2;

        if (start <= root->start && end >= root->end) {

            return root->count;

        }

        else if (mid < start) {

            return query(root->right, start, end);

        }

        else if (mid + 1 > end) {

            return query(root->left, start, end);

        }

        else {

            return query(root->left, start, mid) + query(root->right, mid + 1, end);

        }

    }

};