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.

Example

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

Solution

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的赋值：取左右子树的max的较大值，最后一层的独立结点的max为对应数组中的值。

Solution

public 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;
    }
}