Problem
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.
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]
Challenge
Do it in O(h) time, h is the height of the segment tree.
Note
和segment tree其它题目一样,依然是递归的做法。注意:若树的结点范围为0~n
,那么至少有n/2
个数在左子树上,所以if
语句里用了<=
号。另外,最后一句root.max = Math.max(root.left.max, root.right.max)
是调用递归之后的结果,必须写在最后面。
Solution
public class Solution {
public void modify(SegmentTreeNode root, int index, int value) {
if (index > root.end || index < root.start) return;
if (root.start == root.end) {
root.max = value;
return;
}
if (index <= (root.start + root.end) / 2) modify(root.left, index, value);
else modify(root.right, index, value);
root.max = Math.max(root.left.max, root.right.max);
}
}
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