Lintcode: Segment Tree Query

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.

Have you met this question in a real interview? Yes
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

 

這道題的啓示是：Segment Tree 不光是能夠找線段和，還能夠找線段max, min等各類各樣的指標

 1 /**
 2  * Definition of SegmentTreeNode:
 3  * public class SegmentTreeNode {
 4  *     public int start, end, max;
 5  *     public SegmentTreeNode left, right;
 6  *     public SegmentTreeNode(int start, int end, int max) {
 7  *         this.start = start;
 8  *         this.end = end;
 9  *         this.max = max
10  *         this.left = this.right = null;
11  *     }
12  * }
13  */
14 public class Solution {
15     /**
16      *@param root, start, end: The root of segment tree and 
17      *                         an segment / interval
18      *@return: The maximum number in the interval [start, end]
19      */
20     public int query(SegmentTreeNode root, int start, int end) {
21         // write your code here
22         if (root.start==start && root.end==end) return root.max;
23         int mid = (root.start+root.end)/2;
24         if (end <= mid) return query(root.left, start, end);
25         else if (start > mid) return query(root.right, start, end);
26         else return Math.max(query(root.left, start, mid), query(root.right, mid+1, end));
27     }
28 }