## Bankbazaar Interview Question

Country: India
Interview Type: Phone Interview

Comment hidden because of low score. Click to expand.
1
of 1 vote

One way of doing it is to traverse the nodes in any order (preorder for example), and for every node, traverse it's left subtree and right subtree, for every node in left subtree such that it's value is greater than the current node value, we output a pair, same for right subtree when we find values lesser than the current node value.

``````class Node {
int value;
Node left;
Node right;
}

int ans = 0;

void traverseLeft(Node node, int value){
if (node == null){
return;
}

if (node.value > value){
ans++;
}

traverseLeft(node.left, value);
traverseLeft(node.right, value);
}

void traverseRight(Node node, int value){
if (node == null){
return;
}

if (node.value < value){
ans++;
}

traverseRight(node.left, value);
traverseRight(node.right, value);
}

void traverse(Node node){
if (node == null){
return;
}

traverseLeft(node.left, node.value);
traverseRight(node.right, node.value);

traverse(node.left);
traverse(node.right);
}

traverse(root);``````

Comment hidden because of low score. Click to expand.
-1
of 1 vote

Traverse the binary tree inOrder.
Then run margesort for flip inversion count.

Name:

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