max height of binary search tree

Question

so I need to find the max height of a binary tree but for some reason the result of the code provided below is off by 1. For example if the max height is 3 the following code will give me 2. if the max height is 4 the result will be 3. I am not sure why? the root is not considered for the calculations of the max height therefore i set leftCounter and rightCounter to be 0. any ideas?

public int getMaxHeight(BST.TreeNode<E> n) {
    if(n == null) {
        return 0;
    }
    int leftCounter = 0;
    int rightCounter = 0;
    if(n.left != null) {
        leftCounter = getMaxHeight(n.left) +1 ;
    }
    if(n.right != null) {
        rightCounter = getMaxHeight(n.right) +1 ;
    }

    if(leftCounter > rightCounter) {
        return leftCounter;
    }   
    else 
        return rightCounter;    

}

the max height of this binary tree should be 3 : because of the elements 5,9,11. the root is not counted for the max height.

        15
   _____|____
   10       21 
___|___    __|__ 
9    14   16   24
                |__
                  25

Answer 1

Your code actually returns the correct value; it's just that you misunderstand what the height of a tree means. The height is the number of edges on the longest path from root to leaf, not the number of nodes on the path. So the following tree

        3
   _____|____
   4         5 
___|___    __|__ 
6     7    8   9

has a height of 2, not 3. What you are looking for is the number of levels in the tree, not the height.

public int getNumberOfLevels(BST.TreeNode<E> n) {
    if(n == null) return 0;
    int left = getNumberOfLevels(n.left);
    int right = getNumberOfLevels(n.right);
    return 1 + Math.max(left, right);
}  

Answer 2

you can initialize leftCounter, and RightCounter by 1. Reason being if(n==null) -> false
which means height is atleast 1