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1527896821
tags: Segment Tree, Binary Tree, Divide and Conquer, DFS, Lint
#### Segment Tree
- 和 Segment Tree Query I 以及其他Segment Tree类似: 这个SegmentTreeNode return count of elements in range
- 这个题目考了validate input source：input 的start,end可能超出root[start,end]。
- 那么第一步就要先clear一下: 1. 完全不在range就return 0. 2. 有range重合就规整到root的range.
```
/*
For an array, we can build a SegmentTree for it, each node stores an extra attribute count to
denote the number of elements in the the array which value is between interval start and end.
(The array may not fully filled by elements)
Design a query method with three parameters root, start and end,
find the number of elements in the in array's interval [start, end] by the given root of value SegmentTree.
Example
For array [0, 2, 3], the corresponding value Segment Tree is:
[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
Note
It is much easier to understand this problem if you finished Segment Tree Buildand Segment Tree Query first.
Tags Expand
LintCode Copyright Binary Tree Segment Tree
*/
/*
Thoughts:
Since SegmentTree is already constructed, just use it to calculate count.
If all on left/right, easy, just return that portion.
If mid is between start,end, deal with it carefully.
*/
/**
* Definition of SegmentTreeNode:
* public class SegmentTreeNode {
* public int start, end, count;
* public SegmentTreeNode left, right;
* public SegmentTreeNode(int start, int end, int count) {
* this.start = start;
* this.end = end;
* this.count = count;
* this.left = this.right = null;
* }
* }
*/public class Solution {
public int query(SegmentTreeNode root, int start, int end) {
if (root == null || start > end) {
return 0;
}
if (root.start == start && root.end == end) {
return root.count;
}
//Validation1: Check if out range. If so, set border to root[start,end]
if ((start < root.start && end < root.start) ||
(start > root.end && end > root.end)) {
return 0;
}
// Validate2: if partially overlap, set correct border
if (start < root.start) {
start = root.start;
}
if (end > root.end) {
end = root.end;
}
int mid = (root.start + root.end)/2;
if (end <= mid) {
return query(root.left, start, end);
}
if (start > mid) {
return query(root.right, start, end);
}
//mid in between [start, end]
return query(root.left, start, root.left.end) + query(root.right, root.right.start, end);
}
}
```