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);
}
}
};