Segment Tree Build

36% Accepted

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.

Have you met this question in a real interview? Yes
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

Tags Expand

• LintCode Copyright
• Binary Tree
• Segment Tree

public class Solution {
    /**
     *@param start, end: Denote an segment / interval
     *@return: The root of Segment Tree
     */
    public SegmentTreeNode build(int start, int end) {
        // write your code here
        if (start > end) {
            return null;
        }
        if (start == end) {
            return new SegmentTreeNode(start, end);
        }
        SegmentTreeNode root = new SegmentTreeNode(start, end);
        root.left = build(start, (start + end) /2);
        root.right = build((start + end) / 2 + 1, end);

        return root;
    }
}