［LintCode］Segment Tree Query 线段树的查询

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
It is much easier to understand this problem if you finished Segment Tree Build first.

/**
 * Definition of SegmentTreeNode:
 * public class SegmentTreeNode {
 *     public int start, end, max;
 *     public SegmentTreeNode left, right;
 *     public SegmentTreeNode(int start, int end, int max) {
 *         this.start = start;
 *         this.end = end;
 *         this.max = max
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     *@param root, start, end: The root of segment tree and 
     *                         an segment / interval
     *@return: The maximum number in the interval [start, end]
     */
    public int query(SegmentTreeNode root, int start, int end) {
        if(null == root) return Integer.MIN_VALUE;
        if(root.start > end || root.end < start || start > end) return Integer.MIN_VALUE;
        if(root.start >= start && root.end <= end) return root.max;

        int mid = root.start + (root.end - root.start)/2;
        int leftmax = query(root.left, start, Math.min(mid, end));
        int rightmax = query(root.right, Math.max(mid, start), end);
        return Math.max(leftmax, rightmax);
    }
}