資訊專欄INFORMATION COLUMN
摘要：唯一需要注意的就是的賦值取左右子樹的的較大值，最后一層的獨(dú)立結(jié)點(diǎn)的為對應(yīng)數(shù)組中的值。
Segment Tree Build I Problem
The structure of Segment Tree is a binary tree which each node has two attributes start and end denote an segment / interval.
start and end are both integers, they should be assigned in following rules:
The root"s start and end is given by build method.
The left child of node A has start=A.left, end=(A.left + A.right) / 2.
The right child of node A has start=(A.left + A.right) / 2 + 1, end=A.right.
if start equals to end, there will be no children for this node.
Implement a build method with two parameters start and end, so that we can create a corresponding segment tree with every node has the correct start and end value, return the root of this segment tree.
Given start=0, end=3. The segment tree will be:
[0, 3]
/
[0, 1] [2, 3]
/ /
[0, 0] [1, 1] [2, 2] [3, 3]
Given start=1, end=6. The segment tree will be:
[1, 6]
/
[1, 3] [4, 6]
/ /
[1, 2] [3,3] [4, 5] [6,6]
/ /
[1,1] [2,2] [4,4] [5,5]
Clarification
Segment Tree (a.k.a Interval Tree) is an advanced data structure which can support queries like:
which of these intervals contain a given point
which of these points are in a given interval
See wiki:
Segment Tree
Interval Tree
public class Solution {
public SegmentTreeNode build(int start, int end) {
// write your code here
if (start > end) {
return null;
}
SegmentTreeNode root = new SegmentTreeNode(start, end);
if (start == end) {
return root;
}
root.left = build(start, (start+end)/2);
root.right = build((start+end)/2+1, end);
return root;
}
}
Segment Tree Build II
Difference
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;
}
}
Example
Given [3,2,1,4]. The segment tree will be:
[0, 3] (max = 4)
/
[0, 1] (max = 3) [2, 3] (max = 4)
/ /
[0, 0](max = 3) [1, 1](max = 2)[2, 2](max = 1) [3, 3] (max = 4)
Note
唯一需要注意的就是max的賦值：取左右子樹的max的較大值，最后一層的獨(dú)立結(jié)點(diǎn)的max為對應(yīng)數(shù)組中的值。
Solutionpublic class Solution {
public SegmentTreeNode build(int[] A) {
// write your code here
return build(A, 0, A.length - 1);
}
public SegmentTreeNode build(int[] A, int start, int end) {
if (start > end) {
return null;
}
SegmentTreeNode root = new SegmentTreeNode(start, end, Integer.MIN_VALUE);
if (start != end) {
int mid = (start + end) / 2;
root.left = build(A, start, mid);
root.right = build(A, mid+1, end);
root.max = Math.max(root.left.max, root.right.max);
}
else root.max = A[end];
return root;
}
}
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