Range Sum Query 2D – Mutable

  • My natural naive knowledge is:
    • either have update() in O(1) time and sumRange() in O(n) time: have a loop to calculate the sum each time after an update
    • or have update() in O(n) time and sumRange() in O(1) time: have another array to store the sum of all previous elements, thus the sum for f(n) could be deduced from sum(n+1) – sum(n)
  • For sure, my naive approach got TLE.
  • Time to learn a more advanced data structure: Binary Indexed Tree!
    • Leaf nodes are the elements of the input array
    • each internal node represents some merging of the leaf nodes


  • An array representation of tree could be used to represent Segment Trees. For each node at index i, the left child is at index 2*i+1, right child at 2*i+2 and the parent is at floor((i-1)/2).
  • In mathematics and computer science, the floor and ceiling functions map a real number to the largest previous or the smallest following integer, respectively. More precisely, floor(x)  is the largest integer less than or equal to x and ceiling(x)  is the smallest integer greater than or equal to x.
  • There’s a graph here which is very helpful: http://www.geeksforgeeks.org/binary-indexed-tree-or-fenwick-tree-2/
    • Every node has two numbers: one is the index of this node, the other is the value at this node
    • One important formula to get the parent index of a node at index i: parent(i) = i-i&(-i)
    • The above formula means to remove the right-most bit from i, and then AND with i, and then subtract that result from i, the following code snippet will be helpful, given i = 12, then parent(i) = 8
    • System.out.println(Integer.toBinaryString(12));
      System.out.println(12 – (12&(-12)));1100
    • In the opposite direction, the formula would become parent(i) = i+i&(-i)

Looked at the two posts:

  1. https://discuss.leetcode.com/topic/30343/java-2d-binary-indexed-tree-solution-clean-and-short-17ms
  2. https://discuss.leetcode.com/topic/30935/share-my-java-2-d-binary-indexed-tree-solution

Key notes for me:

  • The biggest time cost is in constructing the binary indexed tree: O(m*n*lgm*lgn), (apparently, there’s a nested fro loop, thus m*n, and inside the nested for loop, it’s calling update(), which is O(lgm*lgn)), however, the tree only needs to be built once, and the benefit this tree brings is that after the tree is built, both update() and sumRange() will have O(lgm*lgn) time.
  • we need an additional array nums[][] to store matrix[][]
  • in the constructor, we build the tree, by calling update(i,j,newVal); function
  • inside update(i,j,newVal); function: we compute the delta between the newVal and the current value stored at nums[i][j], and then
    • store the newVal into this position
    • increment this node and all its parent nodes by delta
      • so, inside this update(i,j,newVal); function, the for loop starts from i+1 and j+1
  • how the sum is computed from the constructed tree:
    • row1,col1 is the overlap that got subtracted twice. That’s why it’s written like this.
int[][] nums;
int[][] tree;
int height;
int width;

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

public void update(int rowIndex, int colIndex, int newVal) {
if(height == 0 || width == 0) return;
int delta = newVal - nums[rowIndex][colIndex];
nums[rowIndex][colIndex] = newVal;
for(int i = rowIndex+1; i <= height; i += i&(-i)){
for(int j = colIndex+1; j <= width; j += j&(-j)){
tree[i][j] += delta;//just use its previous value plus delta is good

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

private 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;


Kth Smallest Element in a BST


A simple in-order traversal could get it AC’ed. Cheers.

[TODO] read others’ more optimal code that didn’t traverse the entire tree, instead, they just return when they found the kth one. Rewrite it myself.

public class Solution {
List<Integer> list;
public int kthSmallest(TreeNode root, int k) {
list = new ArrayList<Integer>();
return list.get(k-1);

void inorder(TreeNode root){
if(root == null) return;
if(root.left != null) inorder(root.left);
if(root.right != null) inorder(root.right);

Recover Binary Search Tree


I attempted to break this one completely by myself, my approach was to find the two nodes by DFS and just swap the two nodes’ values by storing the first value temporarily and DFS the tree three times:

  1. 1st time, to find the first node that should be swapped and store its value in temp1
  2. 2nd time, to find the second node that should be swapped and store its value in temp2, and assign temp1 to be its new value
  3. 3rd time, to find the first node and assign temp1 to be its new value. Done.

Looks like a feasible approach, I even wrote the code for the above approach, but it takes more time to handle the root case (I think I might need process root separately and then do DFS on left and right subtrees.)

Then this most upvoted post amazingly used the simple in-order traversal to have implemented this:


  • inorder traversal is like this (or its template):
    public void inorderTraversal(TreeNode root){inorderTraversal(root.left); //do some business logic;inorderTraversal(root.right)}
  • prev means the one previous to the current root, refer to in-order traversal, previous element must be smaller than the current root, if it’s bigger, then we find the first element, thus we store it in the variable called firstElement
  • prev = root;

    this is the last step in the “do some business logic”, so we’ll always to have update the previous node to be the current root before it traverses the right subtree, since the current root will be the new previous node for the right subtree.

public class Solution {

TreeNode first = null;
TreeNode second = null;
TreeNode prev = new TreeNode(Integer.MIN_VALUE);

public void recoverTree(TreeNode root) {

int temp = first.val;
first.val = second.val;
second.val = temp;

void inorderTraversal(TreeNode root){
if(root == null) return;


if(first == null && prev.val >= root.val){
first = prev;

if(first != null && prev.val >= root.val){
second = root;

prev = root;