timenomad - 1 year ago 78

Java Question

Calculate the longest path between two nodes.

The path is in an arch.

Signature of method is:

`public static int longestPath(Node n)`

In the example binary tree below, it is

This is what I have right now and for the given tree it just returns 0.

`public static int longestPath(Node n) {`

if (n != null) {

longestPath(n, 0);

}

return 0;

}

private static int longestPath(Node n, int prevNodePath) {

if (n != null && n.getLeftSon() != null && n.getRightSon() != null) {

int currNodePath = countLeftNodes(n.getLeftSon()) + countRightNodes(n.getRightSon());

int leftLongestPath = countLeftNodes(n.getLeftSon().getLeftSon()) + countRightNodes(n.getLeftSon().getRightSon());

int rightLongestPath = countLeftNodes(n.getRightSon().getLeftSon()) + countRightNodes(n.getRightSon().getRightSon());

int longestPath = currNodePath > leftLongestPath ? currNodePath : leftLongestPath;

longestPath = longestPath > rightLongestPath ? longestPath : rightLongestPath;

longestPath(n.getLeftSon(), longestPath);

longestPath(n.getRightSon(), longestPath);

return longestPath > prevNodePath ? longestPath : prevNodePath;

}

return 0;

}

private static int countLeftNodes(Node n) {

if (n != null) {

return 1+ countLeftNodes(n.getLeftSon());

}

return 0;

}

private static int countRightNodes(Node n) {

if (n != null) {

return 1+ countRightNodes(n.getRightSon());

}

return 0;

}

I understand that I'm missing a key concept somewhere... My brain goes crazy when I try tracking the flow of execution...

Am I right by saying that by finding the longest path among the root, its left & right nodes and then recurse on its left & right nodes passing them the longest path from previous method invocation and finally (when?) return the longest path, I'm not certain as to how you go about returning it...

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Answer Source

Maybe it is just as simple:

```
public static int longestPath(Node n) {
if (n != null) {
return longestPath(n, 0); // forgot return?
}
return 0;
}
```

Its more complicated than one might think at first sight. Consider the following tree:

```
1
/ \
2 3
/ \
4 5
/ \ \
6 7 8
/ \ \
9 a b
```

In this case, the root node is not even in the longest path (`a-7-4-2-5-8-b`

).

So, what you must do is the following: For each node `n`

you must compute the following:

- compute longest path in left subtree starting with the root of the left subtree (called
`L`

) - compute longest path in right subtree starting with the root of the right subtree (called
`R`

) - compute the longest path in left subtree (not necessarily starting with the root of the left subtree) (called
`l`

) - compute the longest path in right subtree (not necessarily starting with the root of the right subtree) (called
`r`

)

Then, decide, which combination maximizes path length:

`L+R+2`

, i.e. going from a subpath in left subtree to current node and from current node through a subpath in right subtree`l`

, i.e. just take the left subtree and exclude the current node (and thus right subtree) from path`r`

, i.e. just take the right subtree and exclude the current node (and thus left subtree) from path

So I would do a little hack and for every node not return just a single `int`

, but a triple of integers containing `(L+R+2, l, r)`

. The caller then must decide what to do with this result according to the above rules.

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