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.

Example

Example 1:

Input："[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]",1,2
Output：4
Explanation：
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]
The maximum value of [1,2] interval is 4

 思路: 标准模板，必须记住；注意query的区间[start, end]一定是在node的区间范围[node.start, node.end]内的；

/**
 * 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: The root of segment tree.
     * @param start: start value.
     * @param end: end value.
     * @return: The maximum number in the interval [start, end]
     */
    public int query(SegmentTreeNode node, int start, int end) {
        // node.start mid node.end;
        if(node.start == start && node.end == end) {
            return node.max;
        }
        int mid = node.start + (node.end - node.start) / 2;
        int leftmax = Integer.MIN_VALUE;
        int rightmax = Integer.MIN_VALUE;
        // node.start..........mid, mid + 1.........node.end
        //               start              end
        if(start <= mid) {
            leftmax = query(node.left, start, Math.min(end, mid));
        } 
        if(mid + 1 <= end) {
            rightmax = query(node.right, Math.max(start, mid + 1), end);
        }
        return Math.max(leftmax, rightmax);
    }
}