Confused - Height of Binary tree

Question

I'm somewhat confused between the logic of calculating the height of binary tree.

Code 1

public static int findHeight(Tree node) {
    if(node == null)
        return 0;
    else {
        return 1+Math.max(findHeight(node.left), findHeight(node.right));
    }

}

Code 2

public static int findHeight(Tree node) {
    if(node == null)
        return -1;
    else {
        return 1+Math.max(findHeight(node.left), findHeight(node.right));
    }

}

I think, the second one is correct, since it gives the correct answer for below code :-

Tree t4 = new Tree(4);
Tree t2 = new Tree(2);
Tree t1 = new Tree(1);
Tree t3 = new Tree(3);
Tree t5 = new Tree(5);

t4.left = t2;
t4.right = t5;

t2.left = t1;
t2.right = t3;

// Prints "Height : 2" for Code 2
// Prints "Height : 3" for Code 1
System.out.println("Height : " + findHeight(t4)); 

I'm confused because many of other SO answers shows the logic for calculating height as per Code 1

Contradictory logics

• Finding height in Binary Search Tree
• Height of a binary tree

---

UPDATE:

All, I've a basic doubt as to what is exactly the height of tree ?

1. Is it the no of nodes between the root and deepest node of tree ( including both - the root & deepest node ) ?

2. Is it the no of edges between the root and deepest node of tree ?

OR

3. Is it just the matter of implementation of every individual - Both approaches are correct ?

Answer 1

The difference all lies in if an empty tree has height -1 or 0.

According to Wikipedia:

The height of a node is the length of the longest downward path to a leaf from that node. The height of the root is the height of the tree. The depth of a node is the length of the path to its root (i.e., its root path).

and

The root node has depth zero, leaf nodes have height zero, and a tree with only a single node (hence both a root and leaf) has depth and height zero. Conventionally, an empty tree (tree with no nodes, if such are allowed) has depth and height −1.

So you might be right - the second one agrees about this.

Of course, these are all definitions - I would not be too amazed to see a definition that agrees with the first version.