# Java Modified Binary Search Tree Solution

• I understand that the BST is not balanced. The worst case would be O(N^2). Please feel free to give any suggestion or comments. Any one got any idea on coding style? I don't think this lengthy for loop will be a good idea. However, my method for counting min and max are different. Insert is also different.

Basically, the TreeNode maintains 3 additional variables, noOfLeft, noOfRight and noOfSelf. noOfLeft means the how many nodes on the left branch. noOfRight has similar meaning. noOfSelf means how many duplicate nodes.

``````public int countRangeSum(int[] nums, int lower, int upper) {
int len;
if (nums == null || (len = nums.length) <= 0) {
return 0;
} else if (lower > upper) {
return 0;
}

int result = 0;
long sum = nums[0];
if (sum >= lower && sum <= upper) {
result++;
}
TreeNode root = new TreeNode(sum);
for (int i = 1; i < len; i++) {
sum += nums[i];
if (sum >= lower && sum <= upper) {
result++;
}
long max = sum - lower;
long min = sum - upper;

TreeNode node = null;
int count = i;
// Find max
node = root;
while (node != null) {
long value = node.value;
if (max > value) {
node = node.right;
} else if (max < value) {
count -= node.noOfSelf + node.noOfRight;
node = node.left;
} else {
count -= node.noOfRight;
break;
}
}

// Find min
node = root;
while (node != null) {
long value = node.value;
if (min > value) {
count -= node.noOfSelf + node.noOfLeft;
node = node.right;
} else if (min < value) {
node = node.left;
} else {
count -= node.noOfLeft;
break;
}
}

result += count;

// Insert this node.
node = root;
while (node != null) {
long value = node.value;
if (sum > value) {
node.noOfRight++;
if (node.right != null) {
node = node.right;
} else {
node.right = new TreeNode(sum);
break;
}
} else if (sum < value) {
node.noOfLeft++;
if (node.left != null) {
node = node.left;
} else {
node.left = new TreeNode(sum);
break;
}
} else {
node.noOfSelf++;
break;
}
}
}

return result;
}

class TreeNode {
long value;
TreeNode left;
TreeNode right;
int noOfLeft;
int noOfRight;
int noOfSelf;

TreeNode(long value) {
this.value = value;
this.noOfSelf = 1;
}
}``````

• A little refined version with more explanation on variables.

``````public int countRangeSum(int[] nums, int lower, int upper) {
int len;
if (nums == null || (len = nums.length) <= 0) {
return 0;
} else if (lower > upper) {
return 0;
}

int result = 0;
long sum = 0;
TreeNode root = new TreeNode(sum);
for (int i = 0; i < len; i++) {
sum += nums[i];
long max = sum - lower;
long min = sum - upper;

TreeNode node = null;
// Possible combinations.
int count = i + 1;
// Find max
node = root;
while (node != null) {
long value = node.value;
if (max > value) {
node = node.right;
} else if (max < value) {
// Ignore the right parts.
count -= node.noOfSelf + node.noOfRight;
node = node.left;
} else {
count -= node.noOfRight;
break;
}
}

// Find min
node = root;
while (node != null) {
long value = node.value;
if (min > value) {
// Ignore the left parts.
count -= node.noOfSelf + node.noOfLeft;
node = node.right;
} else if (min < value) {
node = node.left;
} else {
count -= node.noOfLeft;
break;
}
}

result += count;

// Insert this node.
node = root;
while (node != null) {
long value = node.value;
if (sum > value) {
node.noOfRight++;
if (node.right != null) {
node = node.right;
} else {
node.right = new TreeNode(sum);
break;
}
} else if (sum < value) {
node.noOfLeft++;
if (node.left != null) {
node = node.left;
} else {
node.left = new TreeNode(sum);
break;
}
} else {
node.noOfSelf++;
break;
}
}
}

return result;
}

class TreeNode {
long value;
TreeNode left;
TreeNode right;
int noOfLeft;
int noOfRight;
int noOfSelf;

TreeNode(long value) {
this.value = value;
this.noOfSelf = 1;
}
}``````

• Balancing trees immediately makes me think of red-black trees, but then it would be tricky to keep track of those noOf counts when performing tree rotations.

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