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Yes, it is ambiguous. In one case it says to print the traversal after each swap operation (which for each T there can be more than 1), then it implies that the traversal only needs to be printed T times.

It would not have taken much effort to make this clear.

Actually, a swap operation consists of more swaps, see the definition for Swapping and for Swap operation, so the problem is not ambiguous when it comes to that part.

How can you assume that the root is 1 and another child is 2? And why are you swapping 1 twice? And what's the meaning of the -1 -1, -1 -1. Makes no sense to me.

I think all the information you need is in the problem description. It's just a lot to take in.

The tree with N nodes is rooted at the node with index 1 as described in the Swapping operation section. So the root of the tree is 1.

-1 is used to represent a null node, as described in the Input Format section. For example, the -1s in the second of the N lines mean that the node with index 2 has NULL pointers for both of its children. Similarly, the -1s in the third of the N lines means that the node with index 3 also has NULL pointers for both of its children.

I don't think there's a particular reason why we swap at level 1 twice. The sample inputs are just used to illustrate an example (Sample Input/Output #00 correspond to Test Case #00 in the Explanation section).

Index 1 and the value of the root are not the same. Binary tree root can contain any value. That part of the problem declaration is ambiguous. For example:

5/ \
32/\/ \
106917

Is still a binary tree. It's not sorted, but that does not disqualify it as a binary tree.

The problem says "You are given a tree of n nodes where nodes are indexed from 1..n and it is rooted at 1. " So we can safely assume that the binary tree we will be given for this problem always has a root of 1. I dealt with this by creating a BinaryTree object with a root of 1.

## Swap Nodes [Algo]

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Yes, it is ambiguous. In one case it says to print the traversal after each swap operation (which for each T there can be more than 1), then it implies that the traversal only needs to be printed T times.

It would not have taken much effort to make this clear.

Actually, a swap operation consists of more swaps, see the definition for

Swappingand forSwap operation, so the problem is not ambiguous when it comes to that part.Input is somehow ambiguous, for example

3

2 3

-1 -1

-1 -1

2

1

1

How can you assume that the root is 1 and another child is 2? And why are you swapping 1 twice? And what's the meaning of the -1 -1, -1 -1. Makes no sense to me.

I think all the information you need is in the problem description. It's just a lot to take in.

The tree with N nodes is rooted at the node with index 1 as described in the Swapping operation section. So the root of the tree is 1.

-1 is used to represent a null node, as described in the Input Format section. For example, the -1s in the second of the N lines mean that the node with index 2 has NULL pointers for both of its children. Similarly, the -1s in the third of the N lines means that the node with index 3 also has NULL pointers for both of its children.

I don't think there's a particular reason why we swap at level 1 twice. The sample inputs are just used to illustrate an example (Sample Input/Output #00 correspond to Test Case #00 in the Explanation section).

Index 1 and the value of the root are not the same. Binary tree root can contain any value. That part of the problem declaration is ambiguous. For example:

Is still a binary tree. It's not sorted, but that does not disqualify it as a binary tree.

The problem says "You are given a tree of n nodes where nodes are indexed from 1..n and it is rooted at 1. " So we can safely assume that the binary tree we will be given for this problem always has a root of 1. I dealt with this by creating a BinaryTree object with a root of 1.