[LintCode] Segment Tree Query I & Segment Tree Query II

Segment Tree Query

Problem

For an integer array (index from 0 to n-1, where n is the size of this array), in the corresponding SegmentTree, each node stores an extra attribute max to denote the maximum number in the interval of the array (index from start to end).

Design a query method with three parameters root, start and end, find the maximum number in the interval [start, end] by the given root of segment tree.

Example

For array [1, 4, 2, 3], the corresponding Segment Tree is:

                  [0, 3, max=4]
                 /             \
          [0,1,max=4]        [2,3,max=3]
          /         \        /         \
   [0,0,max=1] [1,1,max=4] [2,2,max=2], [3,3,max=3]

query(root, 1, 1), return 4

query(root, 1, 2), return 4

query(root, 2, 3), return 3

query(root, 0, 2), return 4

Note

题目是要查询start到end这个区间内某一点的max。max值是从最底层的子节点max值里取最大值。因此，不用太复杂，递归就可以了。

Solution

public class Solution {
    public int query(SegmentTreeNode root, int start, int end) {
        if (root == null || end < root.start || start > root.end) return 0;
        if (root.start == root.end) return root.max;
        return Math.max(query(root.left, start, end), query(root.right, start, end));
    }
}

Segment Tree Query II

Problem

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

与Query I所不同的是，Query II对所给区间内的元素个数求和，而非筛选。这样就会出现start <= root.start && end >= root.end的情况，视作root本身处理。

Solution

public class Solution {
    public int query(SegmentTreeNode root, int start, int end) {
        if (root == null || start > root.end || end < root.start) return 0;
        if (root.start == root.end || (start <= root.start && end >= root.end)) return root.count;
        return query(root.left, start, end) + query(root.right, start, end);
    }
}