Lintcode: Segment Tree Modify

For a Maximum Segment Tree, which each node has an extra value max to store the maximum value in this node's interval.

Implement a modify function with three parameter root, index and value to change the node's value with [start, end] = [index, index] to the new given value. Make sure after this change, every node in segment tree still has the max attribute with the correct value.

Have you met this question in a real interview? Yes
Example
For segment tree:

[1, 4, max=3]
/                \
[1, 2, max=2]                [3, 4, max=3]
/              \             /             \
[1, 1, max=2], [2, 2, max=1], [3, 3, max=0], [4, 4, max=3]
if call modify(root, 2, 4), we can get:

[1, 4, max=4]
/                \
[1, 2, max=4]                [3, 4, max=3]
/              \             /             \
[1, 1, max=2], [2, 2, max=4], [3, 3, max=0], [4, 4, max=3]
or call modify(root, 4, 0), we can get:

[1, 4, max=2]
/                \
[1, 2, max=2]                [3, 4, max=0]
/              \             /             \
[1, 1, max=2], [2, 2, max=1], [3, 3, max=0], [4, 4, max=0]
Note
We suggest you finish problem Segment Tree Build and Segment Tree Query first.

Challenge
Do it in O(h) time, h is the height of the segment tree.

/**
* 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, index, value: The root of segment tree and
*@ change the node's value with [index, index] to the new given value
*@return: void
*/
public void modify(SegmentTreeNode root, int index, int value) {
// write your code here
if (root.start == root.end) {
root.max = value;
return;
}
int mid = (root.start + root.end)/2;
if (index <= mid) modify(root.left, index, value);
else modify(root.right, index, value);
root.max = Math.max(root.left.max, root.right.max);
}
}