LintCode 202 - Segment Tree Query

(202) Segment Tree Query

For an integer array (index from 0 to n-1, where n is the size of this array), in the corresponding SegmentTree, each node stores an extra attributemax to denote the maximum number in the interval of the array (index from start to end).

Design a query method with three parametersroot, start and end, find the maximum number in the interval [start, end] by the given root of segment tree.

Example

For array [1, 4, 2, 3], the corresponding Segment Tree is:

                  [0, 3, max=4]
                 /             \
          [0,1,max=4]        [2,3,max=3]
          /         \        /         \
   [0,0,max=1] [1,1,max=4] [2,2,max=2], [3,3,max=3]

query(root, 1, 1), return 4

query(root, 1, 2), return 4

query(root, 2, 3), return 3

query(root, 0, 2), return 4

    public int query(SegmentTreeNode root, int start, int end) {
        // write your code here
        if (root == null) {
            return Integer.MIN_VALUE;
        }
        if (root.end < start || root.start > end || start > end) {
            return Integer.MIN_VALUE;
        }
        if (start <= root.start && end >= root.end) {
            return root.max;
        }
        int leftWall = root.start + (root.end-root.start)/2;
        int leftMax = query(root.left, start, Math.min(leftWall, end));
        int rightMax = query(root.right, Math.max(leftWall, start), end);
        return Math.max(leftMax, rightMax);
    }

    public int query(SegmentTreeNode root, int start, int end) {

        if (root == null || start > root.end || end < root.start) {

            return 0;

        }

        if (root.start == root.end) {

            return root.max;

        }

        int mid = ((root.start + root.end) / 2);

        return Math.max(query(root.left, start, end), query(root.right, start, end));

    }

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