Find tree height

Problem:
A tree, (NOT NECESSARILY BINARY), has nodes numbered 0 to N-1. An array has indices ranging from 0 to N-1. The indices denote the node ids and values denote the ids of parents. A value of -1 at some index k denotes that node with id k is the root. For ex:

3 3 3 -1 2
0 1 2 3 4

In the above, nodes with ids 0, 1 & 2 have 3 as parent. 3 is the root as its parent = -1 and 2 is the parent of node id 4.

Given such an array, find the height of the tree.

Solution:
A straightforward approach is to find the root, then find all its children, etc. (i.e. perform a BFS), and find out the maximum depth. This is a O(n^2) solution, although it requires only O(1) memory.

Using dynamic programming we can do this in O(n) time using O(n) memory:

public int countDepth(int[] tree) {
    int[] height = new int[tree.length];
    for (int i = 0; i < tree.length; i++)
        height[i] = -1;
    int maxHeight = -1;
    for (int i = 0; i < tree.length; i++) {
        if (height[i] == -1)
            maxHeight = max(maxHeight, findHeight(i, height, tree));
    }
    return maxElement(height);
}

private int findHeight(int i, int[] height, int[] tree) {
    if (height[i] != -1)
        return height[i];
    if (tree[i] == -1) {
        height[i] = 0;
    else
        height[i] = 1 + findHeight(tree[i], height, tree);
    return height[i];
}

Complexity:

time - O(n)
space - O(n)

Links and credits:
http://www.careercup.com/question?id=22809662