Range Sum Query 2D - Mutable

Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

Range Sum Query 2D
The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.

Example:
Given matrix = [
  [3, 0, 1, 4, 2],
  [5, 6, 3, 2, 1],
  [1, 2, 0, 1, 5],
  [4, 1, 0, 1, 7],
  [1, 0, 3, 0, 5]
]

sumRegion(2, 1, 4, 3) -> 8
update(3, 2, 2)
sumRegion(2, 1, 4, 3) -> 10
Note:
The matrix is only modifiable by the update function.
You may assume the number of calls to update and sumRegion function is distributed evenly.
You may assume that row1 ≤ row2 and col1 ≤ col2.

Binary Indexed Tree, O(logm) + O(logn)

public class NumMatrix {

    int[][] tree;
    int[][] nums;
    int m;
    int n;

    public NumMatrix(int[][] matrix) {
        if (matrix.length == 0 || matrix[0].length == 0) return;
        m = matrix.length;
        n = matrix[0].length;
        tree = new int[m+1][n+1];
        nums = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                update(i, j, matrix[i][j]);
            }
        }
    }

    public void update(int row, int col, int val) {
        if (m == 0 || n == 0) return;
        int delta = val - nums[row][col];
        nums[row][col] = val;
        for (int i = row + 1; i <= m; i += i & (-i)) {
            for (int j = col + 1; j <= n; j += j & (-j)) {
                tree[i][j] += delta;
            }
        }
    }

    public int sumRegion(int row1, int col1, int row2, int col2) {
        if (m == 0 || n == 0) return 0;
        return sum(row2+1, col2+1) + sum(row1, col1) - sum(row1, col2+1) - sum(row2+1, col1);
    }

    public int sum(int row, int col) {
        int sum = 0;
        for (int i = row; i > 0; i -= i & (-i)) {
            for (int j = col; j > 0; j -= j & (-j)) {
                sum += tree[i][j];
            }
        }
        return sum;
    }
}

Keep row sum (dp), O(m) + O(n)

public class NumMatrix {
    private int[][] rowSums;
    private int[][] matrix;

    public NumMatrix(int[][] matrix) {
        if(   matrix           == null
           || matrix.length    == 0
           || matrix[0].length == 0   ){
            return;   
         }

         this.matrix = matrix;

         int m   = matrix.length;
         int n   = matrix[0].length;
         rowSums = new int[m][n];
         for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(j == 0)
                    rowSums[i][j] = matrix[i][j];
                else rowSums[i][j] = rowSums[i][j - 1] + matrix[i][j];
            } 
         }
    }
    //time complexity for the worst case scenario: O(m)
    public void update(int row, int col, int val) {
        for(int i = col; i < matrix[0].length; i++){
            rowSums[row][i] = rowSums[row][i] - matrix[row][col] + val;
        }

        matrix[row][col] = val;
    }
    //time complexity for the worst case scenario: O(n)
    public int sumRegion(int row1, int col1, int row2, int col2) {
        int ret = 0;

        for(int j = row1; j <= row2; j++){
            if(col1 == 0) ret += rowSums[j][col2];
            else ret += rowSums[j][col2] - rowSums[j][col1 - 1];
        }

        return ret;
    }
}