线段树查询 II-LintCode

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：

                 [0, 3, count=3]
                 /             \
      [0,1,count=1]             [2,3,count=2]
      /         \               /            \
 [0,0,count=1] [1,1,count=0] [2,2,count=1], [3,3,count=1]

query(1, 1), return 0
query(1, 2), return 1
query(2, 3), return 2
query(0, 2), return 2

#ifndef C247_H
#define C247_H
#include<iostream>
using namespace std;
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 (start > end||root==NULL)
            return 0;
        if (start < root->start)
            return query(root, root->start, end);
        if (end>root->end)
            return query(root, start, root->end);
        SegmentTreeNode *node = root;
        if (start >= node->start&&end <= node->end&&node->count == 0)
            return 0;
        if (start == node->start&&end == node->end)
            return node->count;
        if (end <= node->left->end)
        {
            return query(node->left, start, end);
        }
        else if (start >= node->right->start)
        {
            return query(node->right, start, end);
        }
        else
        {
            int l = query(node->left, start, node->left->end);
            int r = query(node->right, node->right->start, end);
            return l + r;
        }
    }
};
#endif