# Segment Tree Solution in Java

• ``````TreeNode root;
public NumMatrix(int[][] matrix) {
if (matrix.length == 0) {
root = null;
} else {
root = buildTree(matrix, 0, 0, matrix.length-1, matrix[0].length-1);
}
}

public void update(int row, int col, int val) {
update(root, row, col, val);
}

private void update(TreeNode root, int row, int col, int val) {
if (root.row1 == root.row2 && root.row1 == row && root.col1 == root.col2 && root.col1 == col) {
root.sum = val;
return;
}
int rowMid = (root.row1 + root.row2) / 2;
int colMid = (root.col1 + root.col2) / 2;
TreeNode next;
if (row <= rowMid) {
if (col <= colMid) {
next = root.c1;
} else {
next = root.c2;
}
} else {
if (col <= colMid) {
next = root.c3;
} else {
next = root.c4;
}
}
root.sum -= next.sum;
update(next, row, col, val);
root.sum += next.sum;
}

public int sumRegion(int row1, int col1, int row2, int col2) {
return sumRegion(root, row1, col1, row2, col2);
}

private int sumRegion(TreeNode root, int row1, int col1, int row2, int col2) {
if (root.row1 == row1 && root.col1 == col1 && root.row2 == row2 && root.col2 == col2)
return root.sum;
int rowMid = (root.row1 + root.row2) / 2;
int colMid = (root.col1 + root.col2) / 2;
if (rowMid >= row2) {
if (colMid >= col2) {
return sumRegion(root.c1, row1, col1, row2, col2);
} else if (colMid + 1 <= col1) {
return sumRegion(root.c2, row1, col1, row2, col2);
} else {
return sumRegion(root.c1, row1, col1, row2, colMid) + sumRegion(root.c2, row1, colMid+1, row2, col2);
}
} else if (rowMid + 1 <= row1) {
if (colMid >= col2) {
return sumRegion(root.c3, row1, col1, row2, col2);
} else if (colMid + 1 <= col1) {
return sumRegion(root.c4, row1, col1, row2, col2);
} else {
return sumRegion(root.c3, row1, col1, row2, colMid) + sumRegion(root.c4, row1, colMid+1, row2, col2);
}
} else {
if (colMid >= col2) {
return sumRegion(root.c1, row1, col1, rowMid, col2) + sumRegion(root.c3, rowMid+1, col1, row2, col2);
} else if (colMid + 1 <= col1) {
return sumRegion(root.c2, row1, col1, rowMid, col2) + sumRegion(root.c4, rowMid+1, col1, row2, col2);
} else {
return sumRegion(root.c1, row1, col1, rowMid, colMid) + sumRegion(root.c2, row1, colMid+1, rowMid, col2) + sumRegion(root.c3, rowMid+1, col1, row2, colMid) + sumRegion(root.c4, rowMid+1, colMid+1, row2, col2);
}
}
}

private TreeNode buildTree(int[][] matrix, int row1, int col1, int row2, int col2) {
if (row2 < row1 || col2 < col1)
return null;
TreeNode node = new TreeNode(row1, col1, row2, col2);
if (row1 == row2 && col1 == col2) {
node.sum = matrix[row1][col1];
return node;
}
int rowMid = (row1 + row2) / 2;
int colMid = (col1 + col2) / 2;
node.c1 = buildTree(matrix, row1, col1, rowMid, colMid);
node.c2 = buildTree(matrix, row1, colMid+1, rowMid, col2);
node.c3 = buildTree(matrix, rowMid+1, col1, row2, colMid);
node.c4 = buildTree(matrix, rowMid+1, colMid+1, row2, col2);
node.sum += node.c1 != null ? node.c1.sum : 0;
node.sum += node.c2 != null ? node.c2.sum : 0;
node.sum += node.c3 != null ? node.c3.sum : 0;
node.sum += node.c4 != null ? node.c4.sum : 0;
return node;
}

public class TreeNode {
int row1, row2, col1, col2, sum;
TreeNode c1, c2, c3, c4;
public TreeNode (int row1, int col1, int row2, int col2) {
this.row1 = row1;
this.col1 = col1;
this.row2 = row2;
this.col2 = col2;
this.sum = 0;
}
}
``````

Binary Indexed Tree solution is faster and easier compared to this. Just share Segment Tree Solution here. The idea is quite similar to 1D solution. The major difference is that each TreeNode now has 4 children instead of 2.

• @kaiqi it will get 'Exceed Memory Limit" error

• @kaiqi Are we really required to give solutions like this complex in real interviews?

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